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

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

I multiply the integral so that I can get $-x^2$ in the numerator. I then expand the fraction so I can split the integral into easier integrals.

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

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

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

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

I then tried to do the same thing with one of the other integrals I got out of this (the fatal flaw).

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

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

Then some integrals cancel out and all that remains is: $$\int\frac{x^2}{1-x^2+\sqrt{1-x^2}}dx$$

The problem is that this the original integral and I can't think of another way to approach this.

Could someone point out the thing I'm missing? This problem can't be solved with per partes or substitution as I can see.

Solution 1:

The terms $\sqrt{1-x^2}$ motivates us to use change of variable by trigonometric functions. The term $\sqrt{1-x^2}$ naturally implies $x\in[-1,1]$ and so we can assume $x=\sin(t)$, $t\in[-\frac{\pi}{2},\frac{\pi}{2}]$. Then \begin{align*} \int\frac{x^2}{1-x^2+\sqrt{1-x^2}}\,dx&=\int\frac{\sin^2(t)}{\cos^2(t)+\cos(t)}\,\cos(t)\,dt\\ &=\int\frac{1-\cos^2(t)}{1+\cos(t)}\,dt\\ &=\int\left(1-\cos(t)\right)\,dt\\ &=t-\sin(t)+C\\ &=\arcsin(x)-x+C. \end{align*}