How to evaluate the integral: $\int\ln x\;\sin^{-1} x\, \operatorname d\!x$? Can the following integral integrated by parts?
$$\int\ln x\;\sin^{-1} x\, \operatorname d\!x$$
 A: Integration by parts formula
$$\int u(x)\underset{dv}{\underbrace{v^{\prime }(x)dx}}=u(x)v(x)-\int v(x)
\underset{du}{\underbrace{u^{\prime }(x)dx}}.$$
If we apply the LIATE rule to
$$\int \ln x\arcsin x\,dx,$$
we should choose
$$u =\ln x,\quad dv=\arcsin x\,dx.$$
Normally there is a choice of the terms $u,v'$ of the integrand $uv'$ that makes integration much easier. However in the present case the other possible choice $u=\arcsin x,v'=\ln x$ leads to integrals of  similar difficulties. 
$$\begin{eqnarray*}
I &=&\int \ln x\arcsin x\,dx, \qquad \text{(}v =\int \arcsin x\,dx\quad \text{see evaluation below)} \\
 &=&\left( \ln x\right) \left( x\arcsin x+\sqrt{1-x^{2}}\right) -\int \frac{
1}{x}\left( x\arcsin x+\sqrt{1-x^{2}}\right) \,dx \\
&=&\left( \ln x\right) \left( x\arcsin x+\sqrt{1-x^{2}}\right) -\int \arcsin
x\,dx-\int \frac{1}{x}\sqrt{1-x^{2}}\,dx \\
&&\text{(see evaluation of the last integral below)} \\
 &=&\left( \ln x\right) \left( x\arcsin x+\sqrt{1-x^{2}}\right)  -\left( \sqrt{1-x^{2}}+x\arcsin x\right)  \\
&&-\sqrt{1-x^{2}}-\frac{1}{2}\ln \frac{\sqrt{1-x^{2}}-1}{\sqrt{1-x^{2}}+1}+C.
\\
&=&\left( -1+\ln x\right) \left( x\arcsin x+\sqrt{1-x^{2}}\right)  
-\sqrt{
1-x^{2}} -\frac{1}{2}\ln \frac{\sqrt{1-x^{2}}-1}{\sqrt{1-x^{2}}+1}+C.
\end{eqnarray*}$$


*

*Evaluation of $\int \arcsin x\,dx$ by substitution $$\begin{eqnarray*}
\int \arcsin x\,dx &=&\int u\cos u\,du,\quad u=\arcsin x \\
&=&\cos u+u\sin u \\
&=&\sqrt{1-x^{2}}+x\arcsin x+C.
\end{eqnarray*}$$

*Evaluation of $\int \frac{1}{x}\sqrt{1-x^{2}}\,dx$ by substitution and
partial fractions
$$\begin{eqnarray*}
\int \frac{1}{x}\sqrt{1-x^{2}}\,dx &=&-\int \frac{u^{2}}{1-u^{2}}\,du,\qquad
u=\sqrt{1-x^{2}} \\
&=&u+\frac{1}{2}\ln \left( u-1\right) -\frac{1}{2}\ln \left( u+1\right)  \\
&=&u+\frac{1}{2}\ln \frac{u-1}{u+1} \\
&=&\sqrt{1-x^{2}}+\frac{1}{2}\ln \frac{\sqrt{1-x^{2}}-1}{\sqrt{1-x^{2}}+1}+C.
\end{eqnarray*}$$ 

A: Following J.M.'s hint, you may want to prove, again integrating by parts, that
$$\int\log x\,dx=x\log x-x+C\,\,,\,\,C=\,\,\text{a constant}$$and perhaps also to note that
$$\int \frac{x}{\sqrt{1-x^2}}dx=-\frac{1}{2}\int\frac{-2x}{\sqrt{1-x^2}}dx$$
A: Let  I =  ∫ ln x * sin−1 x dx,
= x(ln x - 1) * arcsin x - ∫ x(ln x - 1) * 1/√(1 – x^2) dx
= x(ln x - 1) * arcsin x - ∫(ln x - 1) * x / √(1 – x^2) dx
= x(ln x - 1) * arcsin x + (ln x - 1) * √(1 – x^2) – ∫ (1/x) * √(1 – x^2) dx
Call this last integral  I1. 
Put u = x^2;  then du/dx = 2x
∴ I1 = ∫ (1/(2u)) * √(1 – u) * du/dx dx
Put  u = 1 – v^2, v ≥ 0;  so du/dv = -2v
∴ I1 = ½ ∫ 1/(1 – v^2) * (-2v^2) dv
v^2 / (v^2 – 1) = (v^2 – 1 + 1) / (v^2 – 1) = 1 + 1 / (v^2 – 1)
= 1 + ½ (1/(v – 1)  - 1/(v + 1))
So  I1  = v + ½ ln (A * (v – 1) / (v + 1)),  where  A is the arbitrary constant
= √(1 – u) + ½ ln (A * (√(1 – u) – 1) / (√(1 – u) + 1)),

= √(1 – x^2) + ½ ln (A * (√(1 – x^2) – 1) / (√(1 – x^2) + 1)),

Add this to the other parts, and you are done.
I at first thought it could not be done in elementary functions; but there we are!)
