There is some strategy to solve an integral of this kind?

Question

How to solve the integral

$$\int\frac{\ln x}{\sqrt{1-x}}dx$$

and

$$\int\sqrt{\frac{x}{x-1}}dx$$

I have no idea of how to deal with these integrals.

It's the first integral I attempted.

$$\begin{align} u&=\ln x\\ du&=\frac{dx}{x}\\ \int\frac{\ln x}{\sqrt{1-x}}dx&\stackrel{?}{=}\int\frac{ue^u}{\sqrt{1-e^u}}du \end{align}$$

but only I got something more complicated.

Answer 1

For the first one, integrating by parts: $$ \int \frac{\ln x}{\sqrt{1-x}}\,dx = -2\sqrt{1-x}\ln x+2\int\frac{\sqrt{1-x}}{x}\,dx $$ Then do $u=\sqrt{1-x}$ in the integral, and you will be done after a few simpler steps.

For the second one, let $s=\sqrt{x/(x-1)}$.

Answer 2

For the second integral, use the substitution $$x=\cosh^2\theta$$ Then you can use the formula $$\cosh2\theta=2\cosh^2\theta-1$$ to complete the integration.

Answer 3

For the first integral, do integration by parts with $$u=lnx$$ For the remaining integral, use the substitution $$x=\sin^2\theta$$ You will end up integrating $$2(\csc\theta-\sin\theta)$$

Answer 4

For the first integral, use first integration by parts, setting: $$u=\ln x, \qquad \mathrm d\mkern1mu v=\sqrt{1-x} $$ whence $$\int\frac{\ln x}{\sqrt{1-x}}\mathrm d\mkern1mu x =-2\sqrt{1-x}\,\ln x+2\int\frac{\mathrm d\mkern1mu x}{x\sqrt{1-x}}.$$

Now set $t=\sqrt{1-x}\iff t^2=1-x,\enspace t\ge 0$, whence $\mathrm d\mkern1mu x=-2t\,\mathrm d\mkern1mu t$ and: $$\int\frac{\mathrm d\mkern1mu x}{x\sqrt{1-x}}=-2\int\frac{\mathrm d\mkern1mu t}{1-t^2}=\ln\frac{1-t}{1+t} =\ln\biggl(\frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}\biggr)=2\ln(1-\sqrt{1-x})-\ln x.$$ Finally: $$\int\frac{\ln x}{\sqrt{1-x}}\mathrm d\mkern1mu x =-2(1+\sqrt{1-x})\ln x+4\ln(1-\sqrt{1-x}).$$

For the second integral, it is standard (a function of the square root of a homographic function), you have to set $$u=\sqrt{\dfrac x{1-x}}\iff x=\frac{u^2}{1-u^2}, \quad u\ge 0,$$ so $\,\mathrm d\mkern1mu x= \dfrac{-2u\,\mathrm d\mkern1mu u}{1-u^2}$, and: $$\int\sqrt{\frac{x}{x-1}}\mathrm d\mkern1mu x= \int\dfrac{-2u^2\,\mathrm d\mkern1mu u}{(1-u^2)^2}=2u-2\int\dfrac{\mathrm d\mkern1mu u}{(1-u^2)^2}$$ so you're brought back to the integral of a rational function.