Can someone please help me calculate the following?
$ \int \frac{1+\sqrt{x} }{ \sqrt{x}\sqrt{1-x^2 } } dx $
thanks a lot everyone!
Can someone please help me calculate the following?
$ \int \frac{1+\sqrt{x} }{ \sqrt{x}\sqrt{1-x^2 } } dx $
thanks a lot everyone!
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.
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.
You can split the terms in the numerator. The one with $1$ is a mess, according to Alpha. The one with $\sqrt x$ becomes $\int \frac 1{\sqrt{1-x^2}} dx = \arcsin x$