Question

Integrate:

Answer 1

\[\LARGE \int \frac{dx}{x^4+1}\]

Answer 2

trig sub

Answer 3

x^2= sin x

Answer 4

\[I=\int\limits \frac{ 1 }{ x^4+1 }dx=\frac{ 1 }{ 2 } \int\limits \frac{ 2 }{ x^4+1 }dx=\frac{ 1 }{ 2 }\int\limits \frac{ 2+x^2-x^2 }{ x^4+1 }dx\]
\[=\frac{ 1 }{ 2 }\int\limits \frac{ \left( x^2+1 \right)-\left( x^2-1 \right) }{ x^4+1 }dx\]
\[=\frac{ 1 }{ 2 }\int\limits \frac{ x^2+1 }{ x^4+1 }dx-\frac{ 1 }{ 2 }\int\limits \frac{ x^2-1 }{ x^4+1 }dx\]
\[=\frac{ 1 }{ 2 }I _{1}-\frac{ 1 }{ 2 }I _{2}+c\]
\[I _{1}=\int\limits\frac{ x^2+1 }{ x^4+1 }dx=\int\limits \frac{ 1+\frac{ 1 }{ x^2 } }{ x^2+\frac{ 1 }{ x^2 } }dx\]

Answer 5

xD

Answer 6

for\[I _{1}~put~ x-\frac{ 1 }{ x }=t,\left( 1+\frac{ 1 }{ x^2 } \right)dx=dt\]

Answer 7

\[for ~I _{2},~put~x+\frac{ 1 }{ x }=z,~\left( 1-\frac{ 1 }{ x^2 } \right)dx=dz\]

Answer 8

i think you can proceed further.

Answer 9

how about using

Answer 10

the arctan identity somehow

Answer 11

integral of 1/(x^2+!) = arctac(x)

Answer 12

theres gotta be some simpler way... the first method gets so long

Answer 13

I don't think it'd be valid to try?
\[\LARGE \int \frac{dx}{(x^2)^2+1}\]
And then \(u=x^2\), \(du=2xdx\), \(\sqrt{u}=x\)?

Answer 14

oh u wanna complete squar and do partial frac?

Answer 15

i tried that luigi... u can mess with it, but it wont simply anything,

Answer 16

\[I _{1}=\int\limits \frac{ dt }{ t^2+2 }=\frac{ 1 }{ \sqrt{2} }\tan^{-1} t=\frac{ 1 }{ \sqrt{2} }\tan^{-1} \frac{ x^2-1 }{ x }\]

Answer 17

\[I _{2}=\int\limits \frac{ dz }{ z^2-2 }=\int\limits \frac{ dz }{ \left( z+\sqrt{2} \right)\left( z-\sqrt{2} \right) }=?\]

Answer 18

Alright, thank you guys~