Question

Find $\int \frac{x^4+1}{x(x^2+1{)}^2}$

Answer 1

Given $\int \frac{x^4+1}{x(x^2+1{)}^2}dx$

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

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

$\int \frac{1}{x}dx=ln\left(x\right)$

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

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

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

$=-\frac{1}{2u}$

$=-\frac{1}{2(x^2+1)}$

Putting in solved integrals

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

$=ln\left(x\right)+\frac{1}{x^2+1}+C$