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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.

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2  
Although the question actually begs for a change of variable, you can do it by partial fractions. The trick, of course, is to know how to factor $x^4+1$, which factors as a product of two irreducible quadratics. The roots over $\mathbb{C}$ are $\pm\frac{\sqrt{2}}{2} \pm \frac{\sqrt{2}}{2}i$, so pairing the complex conjugates, you can get that $$x^4 + 1 = (x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1).$$ –  Arturo Magidin Jun 23 '12 at 19:39

3 Answers 3

up vote 4 down vote accepted

Substitute $t = x^2$. This gives us $xdx = \dfrac{dt}2$. Then we get that $$I = \int \dfrac{x}{1+x^4} dx = \dfrac12 \int \dfrac{dt}{1+t^2}$$ Can you take it from here? Move your mouse over the gray area for the answer.

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, $$I = \dfrac12 \int \dfrac{dt}{1+t^2} = \dfrac12 \int \dfrac{\sec^2(\theta) d \theta}{1+\tan^2(\theta)} = \dfrac12 \int \dfrac{\sec^2(\theta) d \theta}{\sec^2(\theta)} \\= \dfrac12 \int d \theta = \dfrac{\theta}2 + C = \dfrac{\arctan(t)}2 + C = \dfrac{\arctan(x^2)}2 + C$$

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Nice exaplanation :) I only knew that from my Integrals Table :) –  Andres Jun 23 '12 at 19:29

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...

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

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

$$I = \frac{1}{2} \arctan (x^2) + C$$

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