Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can someone please help me calculate the following?

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

thanks a lot everyone!

share|improve this question

3 Answers 3

We can express our integral as $$\int \frac{dx}{\sqrt{x}\sqrt{1-x^2}}+\int\frac{dx}{\sqrt{1-x^2}}.$$

The second integral is easy. The first, after the substitution $u=\sqrt{x}$, turns into a constant times $\displaystyle\int \dfrac{du}{\sqrt{1-u^4}}$.

This is a well-known elliptic integral, and cannot be expressed in terms of elementary functions.

share|improve this answer

I will consider the part $\int\frac{1}{\sqrt{x}\sqrt{1-x^2}}dx$ only (The other part is easy). Make the substitution $x=\sin(u)$, we get: $$\int \frac{1}{\sqrt{\sin(u)}\sqrt{1-\sin^2(u)}}\cos(u)du=\int\frac{1}{\sqrt{\sin(u)}}du$$

I believe the last integral is non-elementary.

share|improve this answer

You can split the terms in the numerator. The one with $1$ is a mess, according to Alpha. The one with $\sqrt x$ becomes $\int \frac 1{\sqrt{1-x^2}} dx = \arcsin x$

share|improve this answer
    
thanks a lot ! worked like magic ! –  integrand Dec 29 '12 at 18:16

Your Answer

 
discard

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.