# How can I find $\int\frac1{\sqrt{1+x^4}}\mathrm dx$?

My question is, how can I evaluate the following integral?

$$\int\frac1{\sqrt{1+x^4}}\mathrm dx$$

Thanks.

Amazingly enough, it can be done in terms of elementary functions! I will be sloppy and not worry about details such as making sure that I only take square roots of positive quantities. Throughout, we will try to use only standard'' substitutions (of which there will be quite a few), nothing clever.

Let $x^2=\tan\theta$. So $2x\,dx =\sec^2\theta\,d\theta$. Carry out the substitution as usual, and for the sake of familiarity express the trigonometric functions in terms of $\sin$ and $\cos$. I think we get $$\int\frac{d\theta}{2\cos\theta\sqrt{\sin\theta}}$$ Multiply top and bottom by $\cos\theta$. We get $$\int\frac{\cos\theta\,d\theta}{2(1-\sin^2\theta)\sqrt{\sin\theta}}$$ The natural substitution $y=\sin\theta$ gets us to $$\int\frac{dy}{2(1-y^2)\sqrt{y}}$$ Now let $z=\sqrt{y}$. Then $dy=2z\,dz$, and we end up with $$\int\frac{dz}{1-z^4}$$ Now the game is about over, we use partial fractions as usual.

• @user6312:+1,That's really cute use of substitution :) Apr 1, 2011 at 13:53

Integral

\begin{align*} \int\frac{1}{(1+x^4)^{1/4}}dx&=\int \frac{1}{x(1+1/x^4)^{1/4}}dx\\ &=\int \frac{x^4}{x^5(1+1/x^4)^{1/4}}dx. \end{align*}

Substitution: $z^4=(1+1/x^4)$

$4z^3 dz=-4\frac{1}{x^5}dx$

Therefore, \begin{align*} \int\frac{1}{(1+x^4)^{1/4}}dx&=\int \frac{x^4}{x^5(1+1/x^4)^{1/4}}dx\\ &=-\int\frac{z^2}{(z^4-1)}dz\\ &=-\frac{1}{2}(\frac{1}{z^2-1}+\frac{1}{z^2+1})dz\\ &=-\frac{1}{2}\ln\left|\frac{1-z}{1+z}\right|+\arctan z+ C \end{align*} where $z=(1+1/x^4)^{1/4}$.

• The logarithmic part could be more compactly expressed as an inverse hyperbolic tangent. Dec 4, 2011 at 4:56