A hint to calculate the integral $\int^1_0 \frac{\log x}{1-x}dx$ I'm struggling in trying to solve integral. Could you help me? Please, don't solve the problem, give some hints of what I should do.
$$\int^1_0 \frac{\log x}{1-x}dx$$  
 A: Since the range of the integral is from $0$ to $1$, you may like to use Geometric Series
$$\frac{1}{1-x} = \sum_{k = 0}^{+\infty} x^k$$
Hence
$$ \sum_{k = 0}^{+\infty} \int_0^1 x^k\ln(x)\ dx$$
The integral is pretty standard, you can do it by parts many times or in other ways. Eventually
$$\int x^k\ln(x)\ dx = \frac{x^{k+1} ((k+1) \log (x)-1)}{(k+1)^2}$$
hence
$$\int_0^1 x^k\ln(x) = -\frac{1}{(k+1)^2}$$
This holds for $\Re(k) > -1$ which is true.
Then all together
$$- \sum_{k = 0}^{+\infty} \frac{1}{(k+1)^2} = -\zeta(2) = -\frac{\pi^2}{6}$$
A: This is a non-elementary function integral. Specifically via a simple substitution of $u=1-x \to du = -dx$, we yield the known special function integral : 
$$\int \frac{\ln x}{1-x}\mathrm{d}x=-\int\frac{\ln(1-u)}{u}\mathrm{d}u = \dots =  \text{Li}_2(1-x) + C$$
thus we need to find another way to work around it. 
For an analytical approach, continuing Jack's idea (credits) : 
$$\int x^n\ln(x)\mathrm{d}x = \frac{x^{n+1}\ln(x)}{n+1}-\frac{1}{n+1}\int x^n\mathrm{d}x= \frac{x^{n+1}\big[(n+1)\ln(x)-1\big]}{(n+1)^2} + C$$
for which we used integration by parts as : 
$$\int f\mathrm{d}g = fg - \int g\mathrm{d}f$$
$$\mathrm{d}f = \frac{1}{x}\mathrm{d}x$$
$$g=\frac{x^{n+1}}{n+1}$$
Now, exploiting the well known geometric series :
$$\frac{1}{1-x} = \sum_{k = 0}^{\infty} x^k$$
will lead to the expression for our calculation, of :
$$\sum_{n = 0}^{\infty} \int_0^1 x^n\ln(x)\ dx \to \int_0^1 x^n\ln(x)= -\frac{1}{(n+1)^2} \space \text{for} \space \Re\{n\} > -1$$
which finally leads to the calculation via the usage of the Zeta Function :
$$- \sum_{n = 0}^{\infty} \frac{1}{(n+1)^2} = -\zeta(2) = -\frac{\pi^2}{6}$$
Truly, our calculations are correct, as : 
$$\int_0^1 \frac{\ln x}{1-x}\mathrm{d}x = \text{Li}_2(0) - \text{Li}_2(1) = - \frac{\pi^2}{6}$$
Alternatively, if you're not familiar with such a procedure, the best possible outcome will come via numerical analysis, using Simpson's rule for example : 
$$\int_0^1 \frac{\ln(x)}{1-x}\mathrm{d}x \approx \frac{1-0}{6}\bigg(f(0) +  + 4f\bigg(\frac{1}{2}\bigg) + f(1)\bigg), \space \text{where} \space f(x) = \frac{\ln(x)}{1-x}$$
A: Things to try:


*

*Make the change of variable $x \mapsto 1-x$.  This simplifies the denominator, leaves the interval of integration unchanged (not important, but sometimes useful, especially if a linear combination of the original and new integrands simplifies), and makes the numerator something whose Taylor series you should know.

*Attempt integration by parts to allow you to differentiate the logarithm, turning it into a rational function.  (I'm pretty sure this doesn't help for this problem, but is in the constellation of things to do to integrals that look like this.)

*Try solving the differential equation $f'(x) = \frac{\log x}{1-x}$, which can be turned into a relatively simple recurrence for the Taylor coefficients.

*Look it up in tables.  For instance, in Gradshteyn & Ryzhik, 7th ed., this is 4.231.2.

