7
$\begingroup$

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

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

Thanks.

$\endgroup$
23
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @user6312:+1,That's really cute use of substitution :) $\endgroup$ – Quixotic Apr 1 '11 at 13:53
7
$\begingroup$

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 $z=(1+1/x^4)^{1/4}$.

$\endgroup$
  • $\begingroup$ The logarithmic part could be more compactly expressed as an inverse hyperbolic tangent. $\endgroup$ – J. M. is a poor mathematician Dec 4 '11 at 4:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.