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Can someone please help me calculate the following?

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

thanks a lot everyone!

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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.

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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.

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

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thanks a lot ! worked like magic ! –  integrand Dec 29 '12 at 18:16

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