# Misc integral

$$\int \frac{dx}{x(x^4+1)}$$
$$u=x^2$$
$$\sqrt{u}=x$$
$$dx=\frac{1}{2\sqrt{u}}$$
$$\frac{1}{2}\int \frac{du}{u^2+1}$$
$$\frac{1}{2}arctanx^2+C$$

## Answers and Replies

Mark44
Mentor
$$\int \frac{dx}{x(x^4+1)}$$
$$u=x^2$$
$$\sqrt{u}=x$$
$$dx=\frac{1}{2\sqrt{u}}$$
$$\frac{1}{2}\int \frac{du}{u^2+1}$$
The integral above isn't right. You forgot to replace the x factor in the denominator.
$$\frac{1}{2}arctanx^2+C$$

## The Attempt at a Solution

$$\frac{1}{2} \int \frac{du}{u(u^2+1)}=\int \frac{A}{u}+\frac{Bu+C}{u^2+1}du$$
$$A=\frac{1}{2}=-B$$
$$ln|x|-\frac{1}{4}ln(x^4+1)+C$$