Simple integral with a logarithmic function: $\int\frac{\ln x}{\sqrt{1-x}}\,\mathrm dx$ Consider this indefinite integral and solve it with the simplest way.
$$\int\frac{\ln x}{\sqrt{1-x}}\,\mathrm dx$$
How  can we  solve it?
 A: You can make the substitution $u = 1 - x$ and then apply integration by parts.
Edit:
Using the given substitution, we get (neglecting the negative sign, which you can add in in your work)
$$\int \frac{\ln(1 - u)}{\sqrt{u}}\,\mathrm{d}u = 2\sqrt{u}\ln(1 - u) + 2 \int\frac{\sqrt{u}}{1 - u}\,\mathrm{d}u$$
The last integral can be easily evaluated by putting $u = v^2$.
A: Follow the steps
i) Using integration by parts with $u=\ln(x)$ gives

$$ -2\,\sqrt {1-x}\ln  \left( x \right) -\int \!-2\,{\frac {\sqrt {1-x}}{
x}}{dx}.$$

ii) applying the change of variables $1-x=u^2$ yields

$$ -2\,\sqrt {1-x}\ln  \left( x \right) +4\,\int \!{\frac {{u}^{2}}{{u}^{
2}-1}}{du}
. $$

iii) To evaluate the last integral use the partial fraction technique

$$ \frac{u^2}{u^2-1}=1-\frac{1}{2(1+u)}+ \frac{1}{2(u-1)}. $$

A: Beter make the substitution $z=\sqrt{1-x}$. Then $x=1-z^2$ $dx=-2zdz$
$$\int \frac{\ln (1-z^2)}{z}(-2z dz)=-2\int\ln(1-z^2)dz$$
Now $U=\ln (1-z^2), dU=-\frac{2z}{1-z^2}$,  $dV=dz, V=z$ and an integration by parts gives
$$\int \ln(1-z^2)dz=z\ln (1-z^2)+2\int  \frac{z^2}{1-z^2}dz$$
$$=z\ln (1-z^2)-2\int  \frac{z^2-1+1}{z^2-1}dz$$
$$=z\ln (1-z^2)-2\int dz-2\int \frac{1}{z^2-1}dz=etc.$$
A: Let's take as known the integral
$$\int \ln x\,dx = x\ln x-x+C$$
Start with the substitution $x=1-u^2$, so that $\sqrt{1-x}=u$ and $dx=-2udu$.  We have
$$\int{\ln x\over\sqrt{1-x}}\,dx=-2\int\ln(1-u^2)\,du=-2\int\ln(1-u)\,du-2\int\ln(1+u)\,du$$
Each of the last two integrals can be handled as a straightforward substitution from the known integral of the logarithm function.
Note:  Short of simply differentiating $x\ln x-x$ and verifying you get $\ln x$, I can't offhand think of any way to integrate $\ln x$ without at some point using integration by parts.
