Integral of $\int \frac {1}{1+x \sqrt{1-x^2}} dx$ How to tackle with this problem
$$\int \frac {1}{1+x \sqrt{1-x^2}} dx$$
I have put $x$ = $\sin y$. But couldn't reach at the final result.
Please help....
 A: Hint:
Put $x = 1/y$...
The given integral will transform into
$\frac{-1}{y^2 + \sqrt{y^2-1}} dy$
Rationalize and see what you get...
A: Hint After making the suggested substitution $x = \sin y$, one can transform the resulting integral, which is rational in $\sin y$ and $\cos y$, into a rational function using the Weierstrass substitution $y = 2 \arctan t$. Alternately, one can eliminate the middle step and carry out both substitutions at once, substituting
$$x = \sin (2 \arctan t) = \frac{2t}{1 + t^2}, \quad dx = -\frac{2 (t^2 - 1)}{(t^2 + 1)^2};$$
This substitution looks to produce
$$- 2 \int \frac{t^2 - 1}{t^4 + 2 t^3 + 2 t^2 - 2 t + 1}.$$
The denominator factors as a product of two irreducible quadratic polynomials (but not over $\mathbb{Q}$, some coefficients contain rational multiples of $\sqrt{3}$), and so once (1) one finds these one can (2) proceed with the method of partial fractions, (3) integrate the resulting terms of the form $$\frac{A t + B}{t^2 + C t + D}$$ in the usual way, and (4) back-substitute for the original variable $x$. In particular, the antiderivative is elementary, but consulting a CAS shows that it is somewhat messy.
