Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Integration of $\displaystyle \int\frac{x^2-1}{\sqrt{x^4+1}} \,dx$

$\bf{My\; Try}$:: Let $x^2=\tan \theta$ and $\displaystyle 2xdx = \sec^2 \theta \, d\theta\Rightarrow dx = \frac{\sec^2 \theta}{2\sqrt{\tan \theta}} \, d\theta$

$$ \begin{align} & = \int\frac{\tan \theta - 1}{\sec \theta}\cdot \frac{\sec^2 \theta}{2\sqrt{\tan \theta}} \, d\theta = \frac{1}{2}\int \frac{\left(\tan \theta - 1\right)\cdot \sec \theta}{\sqrt{\tan \theta}} \, d\theta \\ & = \frac{1}{2}\int \left(\sqrt{\tan \theta}-\sqrt{\cot \theta}\right)\cdot \sec \theta \, d\theta \end{align} $$

Now i did not understand how can i solve it

Help me


share|cite|improve this question
Are you sure that this is doable? – imranfat Nov 7 '13 at 16:49
See wolfram. It's not pretty. – amWhy Nov 7 '13 at 16:53
up vote 0 down vote accepted

For any real number of $x$ ,

When $|x|\leq1$ ,





When $|x|\geq1$ ,







share|cite|improve this answer

Binomial expansion of the expression to used in the respective integrals

Use binomial expansions of the expressions in respective integrals as given in the previous solution of the problem.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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