Question

Х dx 4 1 +X

Integrate using partial fractions.

Answer 1

we are given

$\int \frac{x}{1+x^4}dx$

we can also write as

$\int \frac{x}{1+(x^2)^2}dx$

we can use u-subs

LE = n

$du=2xdx$

we get

$=\int \frac{x}{1+u^2}\frac{du}{2x}$

$=\frac{1}{2}\int \frac{1}{1+u^2}du$

$=\frac{1}{2}tan^{-1}(u)+C$

now, we can plug back u=x^2

$=\frac{1}{2}tan^{-1}(x^2)+C$............Answer