Indefinite integral problem. I don't see the trick!

So I have this indefinite integral:

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

My initial hunch is to make $u = 1 + x^4$ but the derivative of that is $4x^3$ but that there is an x in the numerator of the integrand. So I don't see how I can do a u substitution since $x^3 \ne x$. What can I do???!!

• try $u=x^2$.... Commented Apr 14, 2018 at 16:42
• No. It's not so difficult. Try substituting $x^2=u$ (suggested by the $x$ in the numerator). Commented Apr 14, 2018 at 16:42
• The derivative of $\frac{1}{2}\arctan(x^2)$ seems to be... $\frac{x}{1+x^4}$. Commented Apr 14, 2018 at 16:45

Use a substitution $y=x^2.$ Then: $$\int \frac{x}{1+x^4}\, dx=\frac 12\int\frac1{1+y^2}\, dy=\frac{\arctan y}2+c=\frac{\arctan x^2}2+c$$

$$\int \frac{x}{1+x^4} \, dx= \frac12\int \frac{2x}{1+(x^2)^2} \, dx$$

Substitute $u=x^2$ then use the integral of $arctan(x)$.

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

Substitute u = $x^2$ -> $dx = \frac{1}{2x} \, du$

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

The standard integral $\int \frac{1}{u^2+1} \, du = arctan(u)$

$$\frac{1}{2} \int \frac{1}{u^2 + 1} \, du = \frac{arctan(u)}{2}$$

undo substitution $u = x{^2}$

$$= \frac{arctan(x{^2})}{2} + C$$