Integrais Defendidos
editar
nota: ver Integrais
Integrais Defendidos (directos)
editar
nota: directos porque se aplica directamente as formulas da tabela de primitivas.
Passos:
Usar propriedades dos integrais ou da função dada para simplificar o exercício.
Resolver o integral por aplicação directa de uma das formulas.
∫
2
3
x
d
x
{\displaystyle \int \limits _{2}^{3}xdx}
nota: usando
∫
f
(
x
n
)
d
x
=
x
n
+
1
n
+
1
+
C
{\displaystyle \int f(x^{n}){\mbox{d}}x={\frac {x^{n+1}}{n+1}}+C}
n
≠
−
1
{\displaystyle n\neq -1}
∫
2
3
x
d
x
=
[
x
1
+
1
1
+
1
]
2
3
=
[
x
2
2
]
2
3
=
3
2
2
−
2
2
2
=
5
2
{\displaystyle \int \limits _{2}^{3}xdx=\left[{\frac {x^{1+1}}{1+1}}\right]_{2}^{3}=\left[{\frac {x^{2}}{2}}\right]_{2}^{3}={{3^{2}} \over 2}-{{2^{2}} \over 2}={5 \over 2}}
solução:
5
2
{\displaystyle {\frac {5}{2}}}
Integrais definidos por partes
editar
nota: ver Integração por partes .
∫
u
d
v
=
u
v
−
∫
v
d
u
.
{\displaystyle \int u\,dv=uv-\int v\,du.}
∫
e
e
2
x
l
n
(
x
)
d
x
{\displaystyle \int \limits _{e}^{e^{2}}xln(x)dx}
integração por partes, em que
U
=
l
n
(
x
)
,
d
U
=
{\displaystyle U=ln(x),dU=\quad }
1
x
{\displaystyle {1} \over {x}}
d
x
,
V
=
{\displaystyle dx,V=\quad }
x
2
2
{\displaystyle {x^{2}} \over {2}}
d
V
=
x
d
x
{\displaystyle dV=xdx\quad }
∫
x
l
n
(
x
)
d
x
=
l
n
(
x
)
x
2
2
−
∫
x
2
2
∗
1
x
=
x
2
l
n
(
x
)
2
−
1
2
∫
x
2
x
=
x
2
l
n
(
x
)
2
−
1
2
∫
x
=
x
2
l
n
(
x
)
2
−
1
2
∗
x
2
2
=
x
2
l
n
(
x
)
2
−
x
2
4
+
C
{\displaystyle \int xln(x)dx=ln(x){\frac {x^{2}}{2}}-\int {\frac {x^{2}}{2}}*{\frac {1}{x}}={\frac {x^{2}ln(x)}{2}}-{\frac {1}{2}}\int {\frac {x^{2}}{x}}={\frac {x^{2}ln(x)}{2}}-{\frac {1}{2}}\int {x}={\frac {x^{2}ln(x)}{2}}-{\frac {1}{2}}*{\frac {x^{2}}{2}}={\frac {x^{2}ln(x)}{2}}-{\frac {x^{2}}{4}}+C}
logo
[
x
2
l
n
(
x
)
2
−
x
2
4
]
e
e
2
=
e
4
l
n
(
e
2
)
2
−
e
4
4
−
e
2
l
n
(
e
)
2
+
e
2
4
=
e
4
l
n
(
e
)
−
e
2
l
n
(
e
)
2
−
e
4
4
+
e
2
4
=
e
4
l
n
(
e
)
−
2
e
2
l
n
(
e
)
−
e
4
+
e
2
4
{\displaystyle \left[{\frac {x^{2}ln(x)}{2}}-{\frac {x^{2}}{4}}\right]_{e}^{e^{2}}={\frac {e^{4}ln(e^{2})}{2}}-{\frac {e^{4}}{4}}-{\frac {e^{2}ln(e)}{2}}+{\frac {e^{2}}{4}}=e^{4}ln(e)-{\frac {e^{2}ln(e)}{2}}-{\frac {e^{4}}{4}}+{\frac {e^{2}}{4}}=e^{4}ln(e)-{\frac {2e^{2}ln(e)-e^{4}+e^{2}}{4}}}
solução:
e
4
l
n
(
e
)
−
2
e
2
l
n
(
e
)
−
e
4
+
e
2
4
{\displaystyle e^{4}ln(e)-{\frac {2e^{2}ln(e)-e^{4}+e^{2}}{4}}}
F1.2a/21/05/2007
editar
∫
1
2
e
x
l
n
(
x
1
/
5
)
d
x
{\displaystyle \int \limits _{1}^{2e}xln(x^{1/5})dx}
1
5
∫
1
2
e
x
l
n
(
x
)
d
x
{\displaystyle {\frac {1}{5}}\int \limits _{1}^{2e}xln(x)dx}
integração por partes, em que
U
=
l
n
(
x
)
,
d
U
=
{\displaystyle U=ln(x),dU=\quad }
1
x
{\displaystyle {1} \over {x}}
d
x
,
V
=
{\displaystyle dx,V=\quad }
x
2
2
{\displaystyle {x^{2}} \over {2}}
d
V
=
x
d
x
{\displaystyle dV=xdx\quad }
∫
x
l
n
(
x
)
d
x
=
l
n
(
x
)
x
2
2
−
∫
x
2
2
∗
1
x
=
x
2
l
n
(
x
)
2
−
1
2
∫
x
2
x
=
x
2
l
n
(
x
)
2
−
1
2
∫
x
=
x
2
l
n
(
x
)
2
−
1
2
∗
x
2
2
=
x
2
l
n
(
x
)
2
−
x
2
4
+
C
{\displaystyle \int xln(x)dx=ln(x){\frac {x^{2}}{2}}-\int {\frac {x^{2}}{2}}*{\frac {1}{x}}={\frac {x^{2}ln(x)}{2}}-{\frac {1}{2}}\int {\frac {x^{2}}{x}}={\frac {x^{2}ln(x)}{2}}-{\frac {1}{2}}\int {x}={\frac {x^{2}ln(x)}{2}}-{\frac {1}{2}}*{\frac {x^{2}}{2}}={\frac {x^{2}ln(x)}{2}}-{\frac {x^{2}}{4}}+C}
logo
[
x
2
l
n
(
x
)
2
−
x
2
4
]
1
2
e
=
(
2
e
)
2
l
n
(
2
e
)
2
−
(
2
e
)
2
4
−
1
2
l
n
(
1
)
2
+
1
2
4
=
4
e
2
l
n
(
2
e
)
2
−
4
e
2
4
−
l
n
(
1
)
2
+
1
4
=
2
e
2
l
n
(
2
e
)
−
e
2
−
l
n
(
1
)
2
+
1
4
{\displaystyle \left[{\frac {x^{2}ln(x)}{2}}-{\frac {x^{2}}{4}}\right]_{1}^{2e}={\frac {(2e)^{2}ln(2e)}{2}}-{\frac {(2e)^{2}}{4}}-{\frac {1^{2}ln(1)}{2}}+{\frac {1^{2}}{4}}={\frac {4e^{2}ln(2e)}{2}}-{\frac {4e^{2}}{4}}-{\frac {ln(1)}{2}}+{\frac {1}{4}}=2e^{2}ln(2e)-e^{2}-{\frac {ln(1)}{2}}+{\frac {1}{4}}}
então voltamos a atrás e simpliciamos visto ln(1)=0
1
5
(
2
e
2
l
n
(
2
e
)
−
e
2
+
1
4
)
{\displaystyle {\frac {1}{5}}\left(2e^{2}ln(2e)-e^{2}+{\frac {1}{4}}\right)}
solução:
1
5
(
2
e
2
l
n
(
2
e
)
−
e
2
+
1
4
)
{\displaystyle {\frac {1}{5}}\left(2e^{2}ln(2e)-e^{2}+{\frac {1}{4}}\right)}
F1.2b/21/05/2007
editar
∫
−
1
/
2
1
/
2
x
∗
a
r
c
t
g
(
2
x
)
d
x
{\displaystyle \int \limits _{-1/2}^{1/2}x*arctg(2x)dx}
integração por partes, em que
U
=
a
r
c
t
g
(
2
x
)
,
d
U
=
2
1
+
(
2
x
)
2
{\displaystyle U=arctg(2x),dU={\frac {2}{1+(2x)^{2}}}}
d
x
,
V
=
{\displaystyle dx,V=\quad }
x
2
2
{\displaystyle {x^{2}} \over {2}}
d
V
=
x
d
x
{\displaystyle dV=xdx\quad }
∫
x
∗
a
r
c
t
g
(
2
x
)
d
x
=
x
2
a
r
c
t
g
(
2
x
)
2
−
∫
x
2
2
∗
2
1
+
(
2
x
)
2
=
x
2
a
r
c
t
g
(
2
x
)
2
−
∫
x
2
1
+
(
2
x
)
2
+
C
{\displaystyle \int x*arctg(2x)dx={\frac {{x}^{2}\,arctg\left(2\,x\right)}{2}}-\int {\frac {x^{2}}{2}}*{\frac {2}{1+(2x)^{2}}}={\frac {{x}^{2}\,arctg\left(2\,x\right)}{2}}-\int {\frac {x^{2}}{1+(2x)^{2}}}+C}
e usando FOR42/7
∫
x
2
1
+
4
x
2
=
x
4
−
1
4
∫
1
4
x
2
+
1
=
x
4
−
1
4
∗
1
2
a
r
c
t
g
(
2
x
)
=
x
4
−
a
r
c
t
g
(
2
x
)
8
+
C
{\displaystyle \int {\frac {x^{2}}{1+4x^{2}}}={\frac {x}{4}}-{\frac {1}{4}}\int {\frac {1}{4x^{2}+1}}={\frac {x}{4}}-{\frac {1}{4}}*{\frac {1}{2}}arctg(2x)={\frac {x}{4}}-{\frac {arctg\left(2\,x\right)}{8}}+C}
então
∫
x
∗
a
r
c
t
g
(
2
x
)
d
x
=
x
2
a
r
c
t
g
(
2
x
)
2
−
x
4
+
a
r
c
t
g
(
2
x
)
8
=
{\displaystyle \int x*arctg(2x)dx={\frac {{x}^{2}\,arctg\left(2\,x\right)}{2}}-{\frac {x}{4}}+{\frac {arctg\left(2\,x\right)}{8}}=}
x
2
a
r
c
t
g
(
2
x
)
2
+
a
r
c
t
g
(
2
x
)
8
−
x
4
+
C
{\displaystyle {\frac {{x}^{2}\,arctg\left(2\,x\right)}{2}}+{\frac {arctg\left(2\,x\right)}{8}}-{\frac {x}{4}}+C}
logo
[
x
2
a
r
c
t
g
(
2
x
)
2
+
a
r
c
t
g
(
2
x
)
8
−
x
4
]
−
1
/
2
1
/
2
=
1
4
a
r
c
t
g
(
1
)
2
+
a
r
c
t
g
(
1
)
8
−
1
2
4
−
1
4
a
r
c
t
g
(
−
1
)
2
−
a
r
c
t
g
(
−
1
)
8
+
−
1
2
4
=
{\displaystyle \left[{\frac {{x}^{2}\,arctg\left(2\,x\right)}{2}}+{\frac {arctg\left(2\,x\right)}{8}}-{\frac {x}{4}}\right]_{-1/2}^{1/2}={\frac {{\frac {1}{4}}arctg(1)}{2}}+{\frac {arctg(1)}{8}}-{\frac {\frac {1}{2}}{4}}-{\frac {{\frac {1}{4}}arctg(-1)}{2}}-{\frac {arctg(-1)}{8}}+{\frac {-{\frac {1}{2}}}{4}}=}
1
4
a
r
c
t
g
(
1
)
2
+
a
r
c
t
g
(
1
)
8
−
1
4
a
r
c
t
g
(
−
1
)
2
−
a
r
c
t
g
(
−
1
)
8
−
2
8
=
{\displaystyle {\frac {{\frac {1}{4}}arctg(1)}{2}}+{\frac {arctg(1)}{8}}-{\frac {{\frac {1}{4}}arctg(-1)}{2}}-{\frac {arctg(-1)}{8}}-{\frac {2}{8}}=}
π
−
2
8
{\displaystyle {\frac {\pi -2}{8}}}
nota:
a
r
c
t
g
(
1
)
=
π
/
4
{\displaystyle arctg(1)=\pi /4}
a
r
c
t
g
(
−
1
)
=
−
π
/
4
{\displaystyle arctg(-1)=-\pi /4}
solução:
π
−
2
8
{\displaystyle {\frac {\pi -2}{8}}}
Integrais definidos por substituição
editar
nota: ver integração por substituição aqui
∫
1
8
1
x
x
+
1
d
x
{\displaystyle \int \limits _{1}^{8}{1 \over x{\sqrt {x+1}}}dx}
nota: por substituição
t
=
x
+
1
;
x
=
t
2
−
1
;
d
t
=
2
t
{\displaystyle t={\sqrt {x+1}};x=t^{2}-1;dt=2t}
∫
1
x
x
+
1
d
x
=
∫
1
(
t
2
−
1
)
t
2
t
d
t
=
∫
1
(
t
2
−
1
)
2
d
t
=
2
∫
1
(
t
2
−
1
)
d
t
=
2
∫
1
(
t
+
1
)
(
t
−
1
)
d
t
{\displaystyle \int {1 \over x{\sqrt {x+1}}}dx=\int {1 \over (t^{2}-1)t}2tdt=\int {1 \over (t^{2}-1)}2dt=2\int {1 \over (t^{2}-1)}dt=2\int {1 \over (t+1)(t-1)}dt}
usando as propriedades dos logaritmos
l
n
(
x
)
−
l
n
(
y
)
=
l
n
(
x
/
y
)
;
l
n
(
x
)
+
l
n
(
y
)
−
l
n
(
z
)
=
l
n
(
x
y
z
)
{\displaystyle ln(x)-ln(y)=ln(x/y);ln(x)+ln(y)-ln(z)=ln({\frac {xy}{z}})\quad }
2
(
l
n
(
t
−
1
)
2
−
l
n
(
t
+
1
)
2
)
=
l
n
(
t
−
1
)
−
l
n
(
t
+
1
)
{\displaystyle 2({\frac {ln(t-1)}{2}}-{\frac {ln(t+1)}{2}})=ln(t-1)-ln(t+1)\quad }
volta-se atrás na substituição ajustando os indicies à substituição usada; t
[
l
n
(
t
−
1
)
−
l
n
(
t
+
1
)
]
2
3
=
l
n
(
2
)
−
l
n
(
4
)
−
l
n
(
2
−
1
)
+
l
n
(
2
+
1
)
{\displaystyle \left[ln(t-1)-ln(t+1)\right]_{\sqrt {2}}^{3}=ln(2)-ln(4)-ln({\sqrt {2}}-1)+ln({\sqrt {2}}+1)}
solução
l
n
(
2
)
−
l
n
(
4
)
−
l
n
(
2
−
1
)
+
l
n
(
2
+
1
)
{\displaystyle ln(2)-ln(4)-ln({\sqrt {2}}-1)+ln({\sqrt {2}}+1)}
F1.3a/21/05/2007
editar
∫
0
3
[
3
+
(
x
+
1
)
1
/
2
]
/
(
x
+
4
)
d
x
{\displaystyle \int \limits _{0}^{3}\left[3+(x+1)^{1/2}\right]/(x+4)dx}
∫
[
3
+
(
x
+
1
)
1
/
2
]
/
(
x
+
4
)
d
x
{\displaystyle \int \left[3+(x+1)^{1/2}\right]/(x+4)dx}
solução
2
∗
(
−
(
3
∗
a
t
a
n
(
s
q
r
t
(
x
+
1
)
/
s
q
r
t
(
3
)
)
)
/
s
q
r
t
(
3
)
+
(
3
∗
l
o
g
(
x
+
4
)
)
/
2
+
s
q
r
t
(
x
+
1
)
)
{\displaystyle 2*(-(3*atan(sqrt(x+1)/sqrt(3)))/sqrt(3)+(3*log(x+4))/2+sqrt(x+1))\quad }
E3a/2/07/2007
editar
∫
0
7
[
3
+
(
x
+
1
)
1
/
3
]
/
(
x
+
4
)
d
x
{\displaystyle \int \limits _{0}^{7}\left[3+(x+1)^{1/3}\right]/(x+4)dx}
∫
[
3
+
(
x
+
1
)
1
/
3
]
/
(
x
+
4
)
d
x
{\displaystyle \int \left[3+(x+1)^{1/3}\right]/(x+4)dx}
solução
3
(
−
a
t
a
n
(
2
(
x
+
1
)
1
3
−
3
1
3
3
5
6
)
3
1
6
+
(
2
3
2
3
+
1
)
l
o
g
(
(
x
+
1
)
2
3
−
3
1
3
(
x
+
1
)
1
3
+
3
2
3
)
2
3
2
3
+
(
3
2
3
−
1
)
l
o
g
(
(
x
+
1
)
1
3
+
3
1
3
)
3
2
3
{\displaystyle 3\,(-{\frac {atan\left({\frac {2\,{\left(x+1\right)}^{\frac {1}{3}}-{3}^{\frac {1}{3}}}{{3}^{\frac {5}{6}}}}\right)}{{3}^{\frac {1}{6}}}}+{\frac {\left(2\,{3}^{\frac {2}{3}}+1\right)\,log\left({\left(x+1\right)}^{\frac {2}{3}}-{3}^{\frac {1}{3}}\,{\left(x+1\right)}^{\frac {1}{3}}+{3}^{\frac {2}{3}}\right)}{2\,{3}^{\frac {2}{3}}}}+{\frac {\left({3}^{\frac {2}{3}}-1\right)\,log\left({\left(x+1\right)}^{\frac {1}{3}}+{3}^{\frac {1}{3}}\right)}{{3}^{\frac {2}{3}}}}}
Integrais Indefinidos
editar
∫
1
x
+
1
d
x
{\displaystyle \int {1 \over {\sqrt {x+1}}}dx}
∫
1
x
+
1
d
x
=
{\displaystyle \int {1 \over {\sqrt {x+1}}}dx=}
∫
(
x
+
1
)
−
1
/
2
d
x
=
{\displaystyle \int (x+1)^{-1/2}dx=}
(
x
+
1
)
−
1
/
2
+
2
/
2
−
1
/
2
+
2
/
2
{\displaystyle {(x+1)^{-1/2+2/2}} \over {-1/2+2/2}}
=
{\displaystyle =}
(
x
+
1
)
1
/
2
1
/
2
{\displaystyle (x+1)^{1/2} \over 1/2}
=
{\displaystyle =}
2
(
x
+
1
)
1
/
2
=
2
x
+
1
+
C
{\displaystyle 2(x+1)^{1/2}=2{\sqrt {x+1}}+C}
solução:
2
x
+
1
+
C
{\displaystyle 2\,{\sqrt {x+1}}+C}
∫
x
l
n
(
x
2
)
d
x
{\displaystyle \int xln(x^{2})dx}
∫
x
l
n
(
x
2
)
d
x
=
2
∫
x
l
n
(
x
)
d
x
{\displaystyle \int xln(x^{2})dx=2\int xln(x)dx}
integração por partes, em que
U
=
l
n
(
x
)
,
d
U
=
{\displaystyle U=ln(x),dU=\quad }
1
x
{\displaystyle {1} \over {x}}
d
x
,
V
=
{\displaystyle dx,V=\quad }
x
2
2
{\displaystyle {x^{2}} \over {2}}
d
V
=
x
d
x
{\displaystyle dV=xdx\quad }
∫
x
l
n
(
x
)
d
x
=
l
n
(
x
)
x
3
2
−
∫
x
2
2
∗
1
x
=
x
2
l
n
(
x
)
2
−
1
2
∫
x
2
x
=
x
2
l
n
(
x
)
2
−
1
2
∫
x
=
x
2
l
n
(
x
)
2
−
1
2
∗
x
2
2
=
x
2
l
n
(
x
)
2
−
x
2
4
+
C
{\displaystyle \int xln(x)dx=ln(x){\frac {x^{3}}{2}}-\int {\frac {x^{2}}{2}}*{\frac {1}{x}}={\frac {x^{2}ln(x)}{2}}-{\frac {1}{2}}\int {\frac {x^{2}}{x}}={\frac {x^{2}ln(x)}{2}}-{\frac {1}{2}}\int {x}={\frac {x^{2}ln(x)}{2}}-{\frac {1}{2}}*{\frac {x^{2}}{2}}={\frac {x^{2}ln(x)}{2}}-{\frac {x^{2}}{4}}+C}
solução:
2
(
x
2
∗
l
n
(
x
)
2
−
x
2
4
)
+
C
{\displaystyle 2({x^{2}*ln(x) \over 2}-{x^{2} \over 4})+C}
∫
x
2
+
1
(
x
2
−
1
)
x
2
{\displaystyle \int {{x^{2}+1} \over {(x^{2}-1)x^{2}}}}
∫
x
2
+
1
(
x
2
−
1
)
x
2
=
{\displaystyle \int {{x^{2}+1} \over {(x^{2}-1)x^{2}}}=}
...
solução:
−
l
n
(
x
+
1
)
+
l
n
(
x
−
1
)
+
{\displaystyle -ln(x+1)+ln(x-1)+\quad }
1
x
{\displaystyle 1 \over x}
F1.1a/21/05/2007
editar
∫
(
x
2
−
1
)
/
(
x
3
−
5
x
2
+
6
x
)
d
x
{\displaystyle \int (x^{2}-1)/(x^{3}-5x^{2}+6x)dx}
nota: usa-se o método de integração de funções racionais .
factorizando obtêm-se:
x
3
−
5
x
2
+
6
x
=
x
∗
(
x
−
3
)
∗
(
x
−
2
)
{\displaystyle x^{3}-5x^{2}+6x=x*(x-3)*(x-2)\quad }
logo
x
2
−
1
x
3
−
5
x
2
+
6
x
=
x
2
−
1
x
(
x
−
3
)
(
x
−
2
)
=
A
x
+
B
x
−
3
+
C
x
−
2
{\displaystyle {\frac {x^{2}-1}{x^{3}-5x^{2}+6x}}={\frac {x^{2}-1}{x(x-3)(x-2)}}={\frac {A}{x}}+{\frac {B}{x-3}}+{\frac {C}{x-2}}\quad }
e igualando ao numerador
A
(
x
−
3
)
(
x
−
2
)
+
B
∗
x
(
x
−
2
)
+
C
∗
x
(
x
−
3
)
=
x
2
−
1
{\displaystyle A(x-3)(x-2)+B*x(x-2)+C*x(x-3)=x^{2}-1\quad }
tem-se para
x
=
0
;
A
(
x
−
3
)
(
x
−
2
)
=
x
2
−
1
⇔
A
=
x
2
−
1
(
x
−
3
)
(
x
−
2
)
⇔
A
=
−
1
6
{\displaystyle x=0;A(x-3)(x-2)=x^{2}-1\Leftrightarrow A={\frac {x^{2}-1}{(x-3)(x-2)}}\Leftrightarrow A=-{\frac {1}{6}}}
x
=
3
;
B
∗
x
(
x
−
2
)
=
x
2
−
1
⇔
B
=
x
2
−
1
x
(
x
−
2
)
⇔
B
=
8
3
{\displaystyle x=3;B*x(x-2)=x^{2}-1\Leftrightarrow B={\frac {x^{2}-1}{x(x-2)}}\Leftrightarrow B={\frac {8}{3}}}
x
=
2
;
C
∗
x
(
x
−
3
)
=
x
2
−
1
⇔
C
=
x
2
−
1
x
(
x
−
3
)
⇔
C
=
−
3
2
{\displaystyle x=2;C*x(x-3)=x^{2}-1\Leftrightarrow C={\frac {x^{2}-1}{x(x-3)}}\Leftrightarrow C=-{\frac {3}{2}}}
∫
(
x
2
−
1
)
/
(
x
3
−
5
x
2
+
6
x
)
d
x
=
−
1
6
∫
1
x
d
x
+
8
3
∫
1
x
−
3
d
x
−
3
2
∫
1
x
−
2
d
x
=
−
l
n
(
x
)
6
+
8
l
n
(
x
−
3
)
3
−
3
l
n
(
x
−
2
)
2
+
C
{\displaystyle \int (x^{2}-1)/(x^{3}-5x^{2}+6x)dx=-{\frac {1}{6}}\int {\frac {1}{x}}dx+{\frac {8}{3}}\int {\frac {1}{x-3}}dx-{\frac {3}{2}}\int {\frac {1}{x-2}}dx=-{\frac {ln\left(x\right)}{6}}+{\frac {8\,ln\left(x-3\right)}{3}}-{\frac {3\,ln\left(x-2\right)}{2}}+C}
solução:
−
l
n
(
x
)
6
+
8
l
n
(
x
−
3
)
3
−
3
l
n
(
x
−
2
)
2
+
C
{\displaystyle -{\frac {ln\left(x\right)}{6}}+{\frac {8\,ln\left(x-3\right)}{3}}-{\frac {3\,ln\left(x-2\right)}{2}}+C}
* F1.1b/21/05/2007
editar
∫
2
x
/
(
1
−
x
2
)
4
/
5
d
x
{\displaystyle \int 2x/(1-x^{2})^{4/5}dx}
∫
2
x
/
(
1
−
x
2
)
4
/
5
d
x
=
2
∫
x
/
(
1
−
x
2
)
4
/
5
d
x
{\displaystyle \int 2x/(1-x^{2})^{4/5}dx=2\int x/(1-x^{2})^{4/5}dx}
solução:
−
5
(
1
−
x
2
)
1
5
{\displaystyle -5\,{\left(1-{x}^{2}\right)}^{\frac {1}{5}}}
* F1.1c/21/05/2007
editar
∫
(
3
x
2
+
2
x
+
9
)
/
[
(
x
−
1
)
2
(
x
+
1
)
]
d
x
{\displaystyle \int (3x^{2}+2x+9)/\left[(x-1)^{2}(x+1)\right]dx}
nota: usa-se o método de integração de funções racionais . ?
(
x
−
1
)
2
(
x
+
1
)
=
x
3
−
x
2
−
x
+
1
{\displaystyle (x-1)^{2}(x+1)=x^{3}-x^{2}-x+1\quad }
solução:
5
∗
l
n
(
x
+
1
)
2
+
l
n
(
x
−
1
)
2
−
7
x
−
1
{\displaystyle {5*ln(x+1) \over 2}+{ln(x-1) \over 2}-{7 \over x-1}}
Integrais Impróprios
editar
nota: ver Integrais Impróprios aqui
* F1.4a/21/05/2007
editar
∫
0
e
l
n
(
x
)
/
(
x
3
+
1
)
d
x
{\displaystyle \int \limits _{0}^{e}ln(x)/(x^{3}+1)dx}
F1.4b/21/05/2007
editar
∫
1
+
∞
2
/
(
x
+
1
)
d
x
{\displaystyle \int \limits _{1}^{+\infty }2/(x+1)dx}
2
lim
b
→
∞
∫
1
b
1
1
+
x
2
d
x
=
2
[
lim
b
→
∞
arctan
b
−
arctan
1
]
=
2
[
π
/
2
−
π
/
4
]
=
π
/
2
{\displaystyle 2\lim _{b\rightarrow \infty }\int _{1}^{b}{\frac {1}{1+x^{2}}}dx=2\left[\lim _{b\rightarrow \infty }\arctan b-\arctan 1\right]=2\left[\pi /2-\pi /4\right]=\pi /2}
O integral improprio converge apenas se o limite converge. Neste caso o integral não converge.
* E5a/2/07/2007
editar
∫
1
1
x
2
l
n
(
x
)
d
x
{\displaystyle \int \limits _{1}^{1}x^{2}ln(x)dx}
x
3
l
o
g
(
x
)
3
−
x
3
9
{\displaystyle {\frac {{x}^{3}\,log\left(x\right)}{3}}-{\frac {{x}^{3}}{9}}}
* E5b/2/07/2007
editar
∫
2
+
∞
e
−
x
(
x
2
+
1
)
d
x
{\displaystyle \int \limits _{2}^{+\infty }e^{-x}(x^{2}+1)dx}
−
(
l
o
g
(
e
)
2
x
2
+
2
l
o
g
(
e
)
x
+
2
)
e
−
l
o
g
(
e
)
x
l
o
g
(
e
)
3
−
1
e
x
l
o
g
(
e
)
{\displaystyle -{\frac {\left({log\left(e\right)}^{2}\,{x}^{2}+2\,log\left(e\right)\,x+2\right)\,{e}^{-log\left(e\right)\,x}}{{log\left(e\right)}^{3}}}-{\frac {1}{{e}^{x}\,log\left(e\right)}}}
funções (x,y): Domínio, Pontos Fronteira, Pontos Interiores e Exteriores
editar
domínio é o conjunto que contém todos os elementos x e y para os quais a função é definida.
Passos:
Determinar o domínio onde a função dada (Z) é valida.
Determinar se e onde os valores que a função toma (x,y) que podem invalidar essa validade.
Desenhar o gráfico da sub-função com especial interesse para os pontos onde (x,y) invalida Z.
Tipos de funções relevantes devido ao seu uso
(
l
n
(
x
)
,
x
,
x
2
)
{\displaystyle (ln(x),{\sqrt {x}},x^{2})}
Domínio = Pontos interiores.
Pontos Exteriores (PE) = espaço onde (x,y) invalida Z. (basicamente tudo que não é domínio).
Pontos fronteira(PF) = pontos limítrofes / conjunto de pontos que delimitam os espaços (D e PE).
Z
(
x
,
y
)
=
l
n
(
16
−
x
2
−
y
2
)
{\displaystyle Z(x,y)=ln(16-x^{2}-y^{2})\quad }
16
−
x
2
−
y
2
>
0
⇔
−
x
2
−
y
2
>
−
16
⇔
x
2
+
y
2
<
16
⇔
x
2
+
y
2
<
4
2
{\displaystyle 16-x^{2}-y^{2}>0\Leftrightarrow -x^{2}-y^{2}>-16\Leftrightarrow x^{2}+y^{2}<16\Leftrightarrow x^{2}+y^{2}<4^{2}}
nota: domínio do ln é maior que 0, ver logaritmo natural .
D
:
Z
(
x
,
y
)
=
{
(
x
,
y
)
∈
R
2
:
x
2
+
y
2
<
16
}
{\displaystyle D:Z(x,y)=\left\{(x,y)\in \mathbb {R} ^{2}:x^{2}+y^{2}<16\right\}}
P
F
:
Z
(
x
,
y
)
=
{
(
x
,
y
)
∈
R
2
:
x
2
+
y
2
=
16
}
{\displaystyle PF:Z(x,y)=\left\{(x,y)\in \mathbb {R} ^{2}:x^{2}+y^{2}=16\right\}}
P
E
:
Z
(
x
,
y
)
=
{
(
x
,
y
)
∈
R
2
:
x
2
+
y
2
>=
16
}
{\displaystyle PE:Z(x,y)=\left\{(x,y)\in \mathbb {R} ^{2}:x^{2}+y^{2}>=16\right\}}
Z
(
x
,
y
)
=
l
n
(
2
x
2
−
4
x
−
1
+
y
)
{\displaystyle Z(x,y)=ln(2x^{2}-4x-1+y)\quad }
2
x
2
−
4
x
−
1
+
y
>
0
⇔
y
>
−
2
x
2
+
4
x
+
1
{\displaystyle 2x^{2}-4x-1+y>0\Leftrightarrow y>-2x^{2}+4x+1}
nota: usando a formula quadrática
−
b
±
b
2
−
4
a
c
2
a
{\displaystyle \textstyle {-b\pm {\sqrt {b^{2}-4ac}} \over 2a}}
y
=
−
4
±
(
−
4
)
2
−
4
∗
(
−
2
)
∗
1
2
∗
(
−
2
)
=
−
4
±
16
+
8
−
4
=
−
4
±
24
−
4
=
−
4
−
4
±
24
−
4
{\displaystyle y={\frac {-4\pm {\sqrt {(-4)^{2}-4*(-2)*1}}}{2*(-2)}}={\frac {-4\pm {\sqrt {16+8}}}{-4}}={\frac {-4\pm {\sqrt {24}}}{-4}}={\frac {-4}{-4}}\pm {\frac {\sqrt {24}}{-4}}}
nota:
24
=
8
∗
3
=
2
3
∗
3
=
2
2
∗
3
=
2
6
{\displaystyle {\sqrt {24}}={\sqrt {8*3}}={\sqrt {2^{3}*3}}=2{\sqrt {2*3}}=2{\sqrt {6}}}
y
=
1
±
2
6
−
4
=
1
±
6
−
2
=
1
∓
6
2
{\displaystyle y=1\pm 2{\frac {\sqrt {6}}{-4}}=1\pm {\frac {\sqrt {6}}{-2}}=1\mp {\frac {\sqrt {6}}{2}}}
logo
y
=
0
{\displaystyle y=0}
x
=
1
∓
6
2
{\displaystyle x=1\mp {\frac {\sqrt {6}}{2}}}
D
:
Z
(
x
,
y
)
=
{
(
x
,
y
)
∈
R
2
:
y
>
−
2
x
2
+
4
x
+
1
}
{\displaystyle D:Z(x,y)=\left\{(x,y)\in \mathbb {R} ^{2}:y>-2x^{2}+4x+1\right\}}
P
F
:
Z
(
x
,
y
)
=
{
(
x
,
y
)
∈
R
2
:
y
=
−
2
x
2
+
4
x
+
1
}
{\displaystyle PF:Z(x,y)=\left\{(x,y)\in \mathbb {R} ^{2}:y=-2x^{2}+4x+1\right\}}
P
E
:
Z
(
x
,
y
)
=
{
(
x
,
y
)
∈
R
2
:
y
≤
−
2
x
2
+
4
x
+
1
}
{\displaystyle PE:Z(x,y)=\left\{(x,y)\in \mathbb {R} ^{2}:y\leq -2x^{2}+4x+1\right\}}
Z
(
x
,
y
)
=
x
y
{\displaystyle Z(x,y)={\sqrt {xy}}}
{\displaystyle {\sqrt {}}}
raiz quadrada .
x
y
≥
0
{\displaystyle xy\geq 0}
D
:
Z
(
x
,
y
)
=
{
(
x
,
y
)
∈
R
2
:
(
x
≥
0
∧
y
≥
0
)
∧
(
x
≤
0
∧
y
≤
0
)
}
{\displaystyle D:Z(x,y)=\left\{(x,y)\in \mathbb {R} ^{2}:(x\geq 0\land y\geq 0)\land (x\leq 0\land y\leq 0)\right\}}
P
F
:
Z
(
x
,
y
)
=
{
(
x
,
y
)
∈
R
2
:
x
=
0
∧
y
=
0
}
{\displaystyle PF:Z(x,y)=\left\{(x,y)\in \mathbb {R} ^{2}:x=0\land y=0\right\}}
P
E
:
Z
(
x
,
y
)
=
{
(
x
,
y
)
∈
R
2
:
(
x
>
0
∧
y
<
0
)
∧
(
x
<
0
∧
y
>
0
)
}
{\displaystyle PE:Z(x,y)=\left\{(x,y)\in \mathbb {R} ^{2}:(x>0\land y<0)\land (x<0\land y>0)\right\}}
Z
(
x
,
y
)
=
x
2
+
y
2
−
9
{\displaystyle Z(x,y)={\sqrt {x^{2}+y^{2}-9}}}
x
2
+
y
2
−
9
≥
0
⇔
x
2
+
y
2
≥
9
⇔
x
2
+
y
2
≥
3
2
{\displaystyle x^{2}+y^{2}-9\geq 0\Leftrightarrow x^{2}+y^{2}\geq 9\Leftrightarrow x^{2}+y^{2}\geq 3^{2}}
nota: circunferência centrada na origem (0,0) e raio 3 (ver formula da circunferência )
D
:
Z
(
x
,
y
)
=
{
(
x
,
y
)
∈
R
2
:
x
2
+
y
2
≥
9
}
{\displaystyle D:Z(x,y)=\left\{(x,y)\in \mathbb {R} ^{2}:x^{2}+y^{2}\geq 9\right\}}
P
F
:
Z
(
x
,
y
)
=
{
(
x
,
y
)
∈
R
2
:
x
2
+
y
2
=
9
}
{\displaystyle PF:Z(x,y)=\left\{(x,y)\in \mathbb {R} ^{2}:x^{2}+y^{2}=9\right\}}
P
E
:
Z
(
x
,
y
)
=
{
(
x
,
y
)
∈
R
2
:
x
2
+
y
2
<
9
}
{\displaystyle PE:Z(x,y)=\left\{(x,y)\in \mathbb {R} ^{2}:x^{2}+y^{2}<9\right\}}
* E1b-1.1-1.2
editar
g
(
x
,
y
)
=
[
(
4
−
x
2
−
y
2
)
1
/
2
]
/
x
2
−
y
2
)
{\displaystyle g(x,y)=[(4-x^{2}-y^{2})^{1/2}]/x^{2}-y^{2})\quad }
Determinar domino, indicar o interior, exterior e fronteira Curvas de Nível
editar
E1a-1.1-1.2-1.3/2/07/2007
editar
f
(
x
,
y
)
=
l
n
(
2
y
−
3
x
+
x
2
−
4
)
{\displaystyle f(x,y)=ln(2y-3x+x^{2}-4)\quad }
Determinar domino, indicar o interior, exterior e fronteira
Determinar equação geral das curvas de nível, represente no domínio as curvas dos pontos f(x;y)=0 e f(x;y)= -1. Provar que não existe limite
editar
E2a/2/07/2007
editar
l
i
m
(
x
−
y
)
/
(
y
1
/
2
−
x
1
/
2
)
{\displaystyle lim(x-y)/(y^{1/2}-x^{1/2})\quad }
(
x
→
0
+
:
y
→
0
+
)
{\displaystyle (x\rightarrow 0^{+}:y\rightarrow 0^{+})}
calcule o limite ou prove que não existe. visto os polinómios terem graus diferentes pode-se afirmar que a função não tem limite.
E2b/2/07/2007
editar
lim
y
→
2
x
→
0
l
i
m
s
i
n
(
x
y
)
/
x
{\displaystyle \lim _{y\rightarrow 2}^{x\rightarrow 0}lim{\mbox{ }}sin(xy)/x\quad }
calcule o limite ou prove que não existe. nota:
lim
z
→
0
s
i
n
(
z
)
z
=
1
{\displaystyle \lim _{z\rightarrow 0}{\frac {sin(z)}{z}}=1}
pela regra da compensação
lim
y
→
2
x
→
0
s
i
n
(
x
y
)
x
y
=
2
{\displaystyle \lim _{y\rightarrow 2}^{x\rightarrow 0}{\frac {sin(xy)}{x}}y=2}
E1c/2/07/2007
editar
lim
y
→
0
x
→
0
(
2
x
2
−
3
y
2
+
x
y
2
x
2
+
y
2
)
{\displaystyle \lim _{y\rightarrow 0}^{x\rightarrow 0}({\frac {2x^{2}-3y^{2}+xy}{2x^{2}+y^{2}}})}
calcule o limite ou prove que não existe.
lim
y
→
m
x
x
→
0
(
2
x
2
−
3
m
2
x
2
+
x
2
m
2
x
2
+
m
2
x
2
)
=
x
2
(
2
−
3
m
2
+
m
)
x
2
(
2
+
m
2
)
=
2
−
3
m
2
+
m
2
+
m
2
{\displaystyle \lim _{y\rightarrow mx}^{x\rightarrow 0}({\frac {2x^{2}-3m^{2}x^{2}+x^{2}m}{2x^{2}+m^{2}x^{2}}})={\frac {x^{2}(2-3m^{2}+m)}{x^{2}(2+m^{2})}}={\frac {2-3m^{2}+m}{2+m^{2}}}}
para recta horizontal; m=0
2
−
0
+
0
1
+
0
=
2
{\displaystyle {\frac {2-0+0}{1+0}}=2}
o limite é 2
para recta a 45º; m=1
2
−
4
1
+
1
=
−
2
2
=
−
1
{\displaystyle {\frac {2-4}{1+1}}=-{\frac {2}{2}}=-1}
o limite é -1
logo não existe limite
Diferencial Total 1ª ordem
editar
Calculo aproximado
editar
* E4a/2/07/2007
editar
4
,
05
3
/
1
,
99
4
{\displaystyle 4,05^{3}/1,99^{4}\quad }
wxMaxima,
125
96059601
{\displaystyle {\frac {125}{96059601}}}
U
(
x
,
y
)
=
x
3
/
y
4
{\displaystyle U(x,y)=x^{3}/y^{4}\quad }
P
0
(
4
;
2
)
;
U
(
P
0
)
=
64
/
16
=
4
{\displaystyle P_{0}(4;2);U(P_{0})=64/16=4\quad }
P
1
(
4
,
05
;
1
,
99
)
;
U
(
P
1
)
=
125
96059601
{\displaystyle P_{1}(4,05;1,99);U(P_{1})={\frac {125}{96059601}}}
Δ
x
=
x
P
1
−
x
P
0
=
0.05
=
5
100
{\displaystyle \Delta _{x}=x_{P_{1}}-x_{P_{0}}=0.05={\frac {5}{100}}}
Δ
y
=
y
P
1
−
y
P
0
=
−
0
,
01
=
−
1
100
{\displaystyle \Delta _{y}=y_{P_{1}}-y_{P_{0}}=-0,01=-{\frac {1}{100}}}
d
U
d
x
=
3
x
2
y
4
{\displaystyle {\frac {dU}{dx}}=3{\frac {x^{2}}{y^{4}}}}
d
U
d
y
=
−
4
x
3
y
5
{\displaystyle {\frac {dU}{dy}}=-{\frac {4x^{3}}{y5}}}
Δ
U
≈
d
U
d
x
Δ
x
+
d
U
d
y
Δ
y
=
3
x
2
y
4
∗
5
100
+
4
x
3
y
5
(
1
100
)
{\displaystyle \Delta _{U}\thickapprox {\frac {dU}{dx}}\Delta _{x}+{\frac {dU}{dy}}\Delta _{y}=3{\frac {x^{2}}{y^{4}}}*{\frac {5}{100}}+{\frac {4x^{3}}{y5}}({\frac {1}{100}})}
Δ
U
P
0
≈
−
3
∗
4
2
2
4
∗
5
100
+
4
∗
4
3
2
5
(
1
100
)
=
23
/
100
{\displaystyle \Delta _{U_{P_{0}}}\thickapprox -{\frac {3*4^{2}}{2^{4}}}*{\frac {5}{100}}+{\frac {4*4^{3}}{2^{5}}}({\frac {1}{100}})=23/100}
4
,
05
3
/
1
,
99
4
≈
U
(
P
0
)
+
Δ
U
P
0
{\displaystyle 4,05^{3}/1,99^{4}\quad \thickapprox U(P_{0})+\Delta _{U_{P_{0}}}}
4
,
05
3
/
1
,
99
4
≈
4
+
23
/
100
=
423
/
100
{\displaystyle 4,05^{3}/1,99^{4}\quad \thickapprox 4+23/100=423/100}
E4b/2/07/2007
editar
e
−
0.06
∗
2
,
95
2
{\displaystyle e^{-0.06}*2,95^{2}\quad }
wxMaxima,
9025
e
−
0.06
{\displaystyle 9025\,{e}^{-0.06}}
U
(
x
,
y
)
=
e
−
x
∗
y
2
{\displaystyle U(x,y)=e^{-x}*y^{2}\quad }
P
0
(
0
;
3
)
;
U
(
P
0
)
=
9
{\displaystyle P_{0}(0;3);U(P_{0})=9\quad }
P
1
(
0
,
06
;
2
,
95
)
;
U
(
P
1
)
=
9025
e
−
0.06
{\displaystyle P_{1}(0,06;2,95);U(P_{1})=9025\,{e}^{-0.06}}
Δ
x
=
x
P
1
−
x
P
0
=
0.06
−
0
=
6
100
{\displaystyle \Delta _{x}=x_{P_{1}}-x_{P_{0}}=0.06-0={\frac {6}{100}}}
Δ
y
=
y
P
1
−
y
P
0
=
2
,
95
−
3
=
−
0.05
=
−
5
100
{\displaystyle \Delta _{y}=y_{P_{1}}-y_{P_{0}}=2,95-3=-0.05=-{\frac {5}{100}}}
d
U
d
x
=
e
−
x
(
−
1
)
y
2
=
−
y
2
e
x
{\displaystyle {\frac {dU}{dx}}=e^{-x}(-1)y^{2}=-{\frac {y^{2}}{e^{x}}}}
d
U
d
y
=
2
y
e
−
x
=
2
y
e
x
{\displaystyle {\frac {dU}{dy}}=2ye^{-x}={\frac {2y}{e^{x}}}}
Δ
U
≈
d
U
d
x
Δ
x
+
d
U
d
y
Δ
y
=
−
y
2
e
x
∗
6
100
+
2
y
e
x
(
−
5
100
)
{\displaystyle \Delta _{U}\thickapprox {\frac {dU}{dx}}\Delta _{x}+{\frac {dU}{dy}}\Delta _{y}=-{\frac {y^{2}}{e^{x}}}*{\frac {6}{100}}+{\frac {2y}{e^{x}}}(-{\frac {5}{100}})}
Δ
U
P
0
≈
−
3
2
e
0
∗
6
100
+
2
∗
3
e
0
(
−
5
100
)
=
−
21
/
25
{\displaystyle \Delta _{U_{P_{0}}}\thickapprox -{\frac {3^{2}}{e^{0}}}*{\frac {6}{100}}+{\frac {2*3}{e^{0}}}(-{\frac {5}{100}})=-21/25}
e
−
0.06
∗
2
,
95
2
≈
U
(
P
0
)
+
Δ
U
P
0
{\displaystyle e^{-0.06}*2,95^{2}\quad \thickapprox U(P_{0})+\Delta _{U_{P_{0}}}}
e
−
0.06
∗
2
,
95
2
≈
9
−
21
/
25
=
204
/
25
{\displaystyle e^{-0.06}*2,95^{2}\quad \thickapprox 9-21/25=204/25}
![{\displaystyle \int \limits _{2}^{3}xdx=\left[{\frac {x^{1+1}}{1+1}}\right]_{2}^{3}=\left[{\frac {x^{2}}{2}}\right]_{2}^{3}={{3^{2}} \over 2}-{{2^{2}} \over 2}={5 \over 2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e24ea7a200b590b045098c92aa9aa777a4cf2bd)
![{\displaystyle \left[{\frac {x^{2}ln(x)}{2}}-{\frac {x^{2}}{4}}\right]_{e}^{e^{2}}={\frac {e^{4}ln(e^{2})}{2}}-{\frac {e^{4}}{4}}-{\frac {e^{2}ln(e)}{2}}+{\frac {e^{2}}{4}}=e^{4}ln(e)-{\frac {e^{2}ln(e)}{2}}-{\frac {e^{4}}{4}}+{\frac {e^{2}}{4}}=e^{4}ln(e)-{\frac {2e^{2}ln(e)-e^{4}+e^{2}}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/19f2857fe86d3991ada74b12effcccac8a2b0a0b)
![{\displaystyle \left[{\frac {x^{2}ln(x)}{2}}-{\frac {x^{2}}{4}}\right]_{1}^{2e}={\frac {(2e)^{2}ln(2e)}{2}}-{\frac {(2e)^{2}}{4}}-{\frac {1^{2}ln(1)}{2}}+{\frac {1^{2}}{4}}={\frac {4e^{2}ln(2e)}{2}}-{\frac {4e^{2}}{4}}-{\frac {ln(1)}{2}}+{\frac {1}{4}}=2e^{2}ln(2e)-e^{2}-{\frac {ln(1)}{2}}+{\frac {1}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5624eeea8866c8ab60fa402cb54318d510346df8)
![{\displaystyle \left[{\frac {{x}^{2}\,arctg\left(2\,x\right)}{2}}+{\frac {arctg\left(2\,x\right)}{8}}-{\frac {x}{4}}\right]_{-1/2}^{1/2}={\frac {{\frac {1}{4}}arctg(1)}{2}}+{\frac {arctg(1)}{8}}-{\frac {\frac {1}{2}}{4}}-{\frac {{\frac {1}{4}}arctg(-1)}{2}}-{\frac {arctg(-1)}{8}}+{\frac {-{\frac {1}{2}}}{4}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fe057ceaee2a60fbd932f89b2304ee34f323aa4)
![{\displaystyle \left[ln(t-1)-ln(t+1)\right]_{\sqrt {2}}^{3}=ln(2)-ln(4)-ln({\sqrt {2}}-1)+ln({\sqrt {2}}+1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ffa463387003ae04fe2f4a514c170c6d81181a9d)
![{\displaystyle \int \limits _{0}^{3}\left[3+(x+1)^{1/2}\right]/(x+4)dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4bc98717a3dcafbe9900baec8b4a2cada807de95)
![{\displaystyle \int \left[3+(x+1)^{1/2}\right]/(x+4)dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98133bdbe901462f027270224fcb7b3d0b4a59d6)
![{\displaystyle \int \limits _{0}^{7}\left[3+(x+1)^{1/3}\right]/(x+4)dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6be034a690cb8dbd78f578d74544f0c9f56e84b7)
![{\displaystyle \int \left[3+(x+1)^{1/3}\right]/(x+4)dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1c5e7e7413b4cd75efdf77e1964d49993651cab)
![{\displaystyle \int (3x^{2}+2x+9)/\left[(x-1)^{2}(x+1)\right]dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be3a15d8089f957c6f38a28149385684157b43ae)
![{\displaystyle 2\lim _{b\rightarrow \infty }\int _{1}^{b}{\frac {1}{1+x^{2}}}dx=2\left[\lim _{b\rightarrow \infty }\arctan b-\arctan 1\right]=2\left[\pi /2-\pi /4\right]=\pi /2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/456448bef4ed8407c6fbe3c10869286777503a96)
![{\displaystyle g(x,y)=[(4-x^{2}-y^{2})^{1/2}]/x^{2}-y^{2})\quad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/20e51280f5af424670e10f3cd0f9e77112c972d7)
