$$ \int\frac{1+x^2}{(1-x^2)(\sqrt{1+x^4})}dx $$

I thought of substituting $ x-\frac{1}{x} $ as $t$ but it gets stuck midway. I am close but I think I need to sustitute something else here.


Let $$I = \int\frac{(1+x^2)}{(1-x^2)\sqrt{1+x^4}}dx = -\int\frac{(1+x^2)}{x^2(x-\frac{1}{x})\sqrt{x^2+\frac{1}{x^2}}}dx$$

So $$I = -\int\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})\sqrt{(x-\frac{1}{x})^2+2}}dx$$

Now Put $\displaystyle x-\frac{1}{x} = t\;,$ Then $\displaystyle \left(1+\frac{1}{x^2}\right)dx = dt$

So $$I = -\int\frac{1}{t\sqrt{t^2+2}}dt$$

Now Put $t = \sqrt{2}\tan \theta\;,$ Then $dt=\sqrt{2}\sec^2 \theta d\theta$

So $$I = -\int\frac{\sqrt{2}\sec^2 \theta}{2\tan \theta \sec \theta}d\theta = -\frac{1}{\sqrt{2}}\int\csc \theta d \theta$$

So $$I = \ln \left|\csc \theta +\cot \theta\right|+\mathcal{C}=\frac{1}{\sqrt{2}}\ln \left|\frac{\sqrt{2}+\sqrt{2+t^2}}{t+\sqrt{2+t^2}}\right|+\mathcal{C}$$

So $$I = \frac{1}{\sqrt{2}}\ln \left|\frac{\sqrt{2}x^2+x\sqrt{1+x^4}}{x^2-1+\sqrt{1+x^4}}\right|+\mathcal{C}$$


Given expression is also equal to: $$\int\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})(\sqrt{\frac{1}{x^2}+x^2})}dx$$Which further reduces to $$\int\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})\sqrt{(x-\frac{1}{x})^2-2}}dx$$Now let $x-\frac{1}{x}=t$ The rest is evident.

  • 1
    $\begingroup$ This is pretty much just what @juantheron did, is it not? $\endgroup$ – S.C.B. May 17 '16 at 13:20
  • $\begingroup$ Well, While writing the answer, I did not know that someone has already posted the answer. $\endgroup$ – Prayas Agrawal May 17 '16 at 15:22

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.