Evaluating an Integral with U-Substitution

[leprofece]

integral ((x+sqrt(1-x^2))/((1-xsqrt(1-x^2))

integral \frac{1 +\sqrt{1-x^2}}{1-x\sqrt{1-x^2}}

I think it can be solved U = 1-x^2
du = -2x
so (-1/2sqrt/(u-1))(1+ u^1/2)/1-sqrt/(u-1)u
it looks I amgoing by the right way am I correct?

 

[Dethrone]

For everyone's benefit, I will rewrite your integral in LaTeX:

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

If this isn't your integral, let me know.

Yes, your are on the right track.

If $u=1-x^2$ => $du=-2xdx$

Rewriting the integral in terms of $u$:

$$=\int \frac{\sqrt{1-u}-\sqrt{u}}{1-\sqrt{1-u}\sqrt{u}}\,du$$
$$=\int \frac{1}{1-\sqrt{u}}\,du-\int \frac{1}{1-\sqrt{1-u}} \,du$$

For the left integral: let $u=s^2$. For the right integral: let $1-u=t^2$
$$=\int \frac{2s}{1-s}\,ds-\int \frac{2t}{1-t} \,dt$$
$$=-2\int \frac{s-1+1}{s-1}\,ds-2\int \frac{t-1+1}{t-1} \,dt$$
$$=-2(\sqrt{u}+\ln\left({\sqrt{u}-1)}+\sqrt{1-u}+\ln\left({\sqrt{1-u}-1}\right)\right) + C$$
$$=-2(\sqrt{1-x^2}+\ln\left({\sqrt{1-x^2}-1)}+\sqrt{x^2}+\ln\left({\sqrt{x^2}-1}\right)\right) + C$$

There should be absolute value signs on the natural logs, and I think you can finish the rest and simplify the answer. If you find a mistake, let me know.