Evaluating $\int{\frac{x}{1+x^4}\ dx}$ I have this integral:
$$\int{\frac{x}{1+x^4}\ dx}$$
The solution should be:
$$\frac{1}{2} \arctan{x^2}+C$$
But I have only seen how to integrate when in denominator I have an expression with real roots. Here, with $1+x^4$ I dont have any real roots and I don't know how I can integrate it. 
 A: Substitute $t = x^2$. This gives us $xdx = \dfrac{dt}2$. Then we get that $$I = \int \dfrac{x}{1+x^4} dx = \dfrac12 \int \dfrac{dt}{1+t^2}$$
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 We get that $\displaystyle I = \dfrac12 \int \dfrac{dt}{1+t^2}$. Now substitute $\theta = \arctan(t)$. This gives us that $t = \tan(\theta)$ and $dt = \sec^2(\theta) d \theta$. Also, note that $1 + \tan^2(\theta) = \sec^2(\theta)$. Hence, $$I = \dfrac12 \int \dfrac{dt}{1+t^2} = \dfrac12 \int \dfrac{\sec^2(\theta) d \theta}{1+\tan^2(\theta)} = \dfrac12 \int \dfrac{\sec^2(\theta) d \theta}{\sec^2(\theta)} \\= \dfrac12 \int d \theta = \dfrac{\theta}2 + C = \dfrac{\arctan(t)}2 + C = \dfrac{\arctan(x^2)}2 + C$$

A: Let $u = x^2$.  Then $\frac{du}{dx} = 2x \Longrightarrow dx = du/2x$.  So we can rewrite the integrand as $\frac{1}{2} (\frac{du}{1 + u^2})$.  The rest is a direct trig identity...
A: $$I := \int \frac{x}{1+x^4} \ dx$$
Let $u = x^2.$ Then $du = 2x \ dx$. So, we now have:
$$I = \frac{1}{2} \int \frac{du}{1+u^2}$$
Recall that $$\int \frac{du}{1+u^2} = \arctan (u) +C$$
So, we now have that
$$I = \frac{1}{2} \arctan (u) + C$$
Replacing $u$ with $x^2$ leads us to the desired result of
$$I = \frac{1}{2} \arctan (x^2) + C$$
