10
$\begingroup$

Question. To find the integral of- $$\int {\frac{1}{(x^4+1)^\frac{1}{4}} \, \mathrm{d}x}$$

I have tried substituting $x^4+1$ as $t$, and as $t^4$, but it gives me an even more complex integral. Any help?

$\endgroup$
2
  • $\begingroup$ Wolfram gives me a difficult result, but it is in the form of $arctan$ and $ln$ $\endgroup$
    – Athena
    Jun 30, 2016 at 9:20
  • $\begingroup$ @b00nheT, I am not familiar, regrettably, with hyperbolic functions. It isn't in our syllabus, either. We are to find it without those functions. $\endgroup$
    – Athena
    Jun 30, 2016 at 9:23

2 Answers 2

19
$\begingroup$

Hint:

This can be written as : $$\int \frac{x^4dx}{x^5\left(1+\frac{1}{x^4}\right)^{1/4}}$$ Now substitute $1+\frac{1}{x^4}=t^4$ $$\implies t^3dt=-\frac{1}{x^5}dx$$ and $$x^4=\frac{1}{t^4-1}$$ to get $$\int \frac{t^2dt}{1-t^4}$$ Now use partial fractions.

$\endgroup$
1
  • 2
    $\begingroup$ Just when I figured it out. Exactly one moment before. Thanks, mate. $\endgroup$
    – Athena
    Jun 30, 2016 at 9:32
10
$\begingroup$

Let $$I = \int\frac{1}{(x^4+1)^{\frac{1}{4}}}dx$$

Put $x^2=\tan \theta,$ Then $2xdx = \sec^2 \theta d\theta$

So $$I = \int\frac{\sec^2 \theta}{\sqrt{\sec \theta}}\cdot \frac{1}{2\sqrt{\tan \theta}}d\theta = \frac{1}{2}\int\frac{1}{\cos \theta \sqrt{\sin \theta}}d\theta = \frac{1}{2}\int\frac{\cos \theta}{(1-\sin^2 \theta)\sqrt{\sin \theta}}d\theta$$

Now Put $\sin \theta = t^2\;,$ Then $\cos \theta d\theta = 2tdt$

So $$I = \int\frac{1}{1-t^4}dt = -\int\frac{1}{(t^2-1)(t^2+1)}dt = -\frac{1}{2}\int\left[\frac{1}{1-t^2}+\frac{1}{1+t^2}\right]dt$$

So $$I = \frac{1}{2}\ln \left|\frac{t-1}{t+1}\right|-\frac{1}{2}\tan^{-1}(t)+\mathcal{C}$$

$\endgroup$
2
  • 3
    $\begingroup$ Nice!(+1) but the $tan^{-1}t$ is missing a coefficient of $\frac{1}{2}$, otherwise our answers match. $\endgroup$
    – Nikunj
    Jun 30, 2016 at 11:47
  • $\begingroup$ Thanks Nikung, But your,s solution is much better then mine. $\endgroup$
    – juantheron
    Jun 30, 2016 at 12:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .