Indefinite integration : $\int \frac{1+x-x^2}{\sqrt{(1-x^2)^3}}$ Problem : 
Solve : $\int \frac{1+x-x^2}{\sqrt{(1-x^2)^3}}$
I tried : 
$\frac{1-x^2}{\sqrt{(1-x^2)^3}} + \frac{x}{\sqrt{(1-x^2)^3}}$
But it's not working....Please guide how to proceed 
 A: Hint:when $|x|\lt 1$
$$\int(\frac{1-x^2}{\sqrt{(1-x^2)^3}} + \frac{x}{\sqrt{(1-x^2)^3}})dx=\int\frac{dx}{\sqrt{(1-x^2)}} + \frac{-1}{2}\int\frac{-2xdx}{\sqrt{(1-x^2)^3}}$$ take $$u=1-x^2$$
A: What you did is a very good first step. 
For the second integral, make the substitution $u=1-x^2$.  We end up with the easy $\int -\frac{1}{2}u^{-3/2}\,du $. 
For the first integral, note that the bottom simplifies to $(1-x^2)\sqrt{1-x^2}$, since the square root is only defined when $|x|\le 1$. There is nice cancellation, and we are integrating $\frac{1}{\sqrt{1-x^2}}$, easy, we get an $\arcsin$.
A: It will work the way you have split the integrand. The first term you can simplify to $\int{\frac{1}{\sqrt{1-x^2}}}dx$ and in the second term you can make $x^2=y$. 
A: you had a good guees! see
$$\int \frac{1+x-x^2}{(1-x^2)^{\frac{3}{2}}}dx = \int \frac{1-x^2}{(1-x^2)^{\frac{3}{2}}}+\frac{x}{(1-x^2)^{\frac{3}{2}}}dx
=\int (1-x^2)^{\frac{-1}{2}}dx+\int\frac{x}{(1-x^2)^{\frac{3}{2}}}dx$$
note that $[(1-x^2)^{\frac{-1}{2}}] '= -\frac{1}{2}(1-x^2)^\frac{-3}{2}(-2x)= \frac{x}{(1-x)^{\frac{3}{2}}}$ . Now let's us solve changing the variable $x= \sin t$ then $dx=\cos tdt$ therefore
$$\int (1-x^2)^{\frac{-1}{2}}dx=\int (1-\sin^2t)^{\frac{-1}{2}}\cos t \;dt=
\int \frac{\cos t}{cost} dt=\int dt=t =\arcsin x$$
So we get
$$\int \frac{1+x-x^2}{(1-x^2)^{\frac{3}{2}}}dx=\int (1-x^2)^{\frac{-1}{2}}dx+\int\frac{x}{(1-x^2)^{\frac{3}{2}}}dx=\frac{x}{(1-x)^{\frac{3}{2}}}+\arcsin x$$
A: Using math_man's substitution throughout, $$x=\sin(u)$$So: $$(1-x^2)^\frac{1}{2}=\cos(u)$$and$$dx=\cos(u) du$$
The integral becomes: $$\int \frac{(\cos^2(u)+\sin(u))}{\cos^3(u)}\cos(u) du=\int du+\int \frac{\sin(u)}{\cos^2(u)}du$$Both these integrals are simple.
