Find an integral splitting it into the power series I want to find an integral splitting it into the power series. 
$$\int_0^1\ln\frac{1}{1 - x}dx$$
I tried to split $\ln$ into Maclaurin series:
$$\int_0^1\ln\frac{1 - x + x}{1 - x}dx  = \int_0^1\ln\left(1 + \frac{x}{1 - x}\right)dx =\int_0^1\left(\sum_{k = 0}^{\infty}\frac{x^k}{k(1 - x)^k}\right)dx$$
But seems it didn't help me. What can I try to evaluate this expression ?
 A: Unless you're forced to use power series, I don't think that power series are the best way to solve this problem.  However, if you really want to use power series, note that we can begin by simplifying the integrand using the laws of logarithms to get
$$
\int_0^1\ln\frac{1}{1-x}dx=-\int_0^1\ln(1-x)dx.
$$
Next, observe that this integral is an improper integral since $\ln(1-1)=\ln(0)$ which is undefined.  Therefore, we rewrite this as
$$
-\lim_{r\rightarrow 1^-}\int_0^r\ln(1-x)dx.
$$
Note that $\ln(1-x)$ has Taylor series expansion $-\sum_{n=1}^\infty\frac{x^n}{n}$ centered at $0$ with radius of convergence $1$.  Therefore, for any $r$ considered above, we can substitute the Taylor expansion to get
$$
\lim_{r\rightarrow 1^-}\int_0^r\sum_{n=1}^\infty\frac{x^n}{n}dx.
$$
Since Taylor (power) series are uniformly convergent within their interval of convergence (this can be made more precise or ignored, depending on your level), we can interchange the limit and the derivative to get that this equals
$$
\lim_{r\rightarrow 1^-}\sum_{n=1}^\infty\int_0^r\frac{x^n}{n}dx=\lim_{r\rightarrow 1^-}\sum_{n=1}^\infty\left.\frac{x^{n+1}}{n(n+1)}\right|_0^r.
$$
Once again, using uniform convergence, we can exchange the sum and the limit to get
$$
\sum_{n=1}^\infty\lim_{r\rightarrow 1^-}\left.\frac{x^{n+1}}{n(n+1)}\right|_0^r=\sum\lim_{r\rightarrow 1^-}\frac{r^{n+1}}{n(n+1)}=\sum_{n=1}^\infty\frac{1}{n(n+1)}.
$$
Finally, we can use a partial fraction decomposition to write this as
$$
\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+1}\right),
$$
which telescopes to $1$. 

Now, if I were solving this without any restrictions, I would take the integral
$$
-\int_0^1\ln(1-x)dx
$$
and use the $u$-substitution $u=1-x$ so $du=-dx$ to get
$$
\int_1^0\ln(u)du.
$$
This integral is, once again, improper when $u=0$, so we write
$$
\lim_{r\rightarrow 0^+}\int_1^r\ln(u)du.
$$
An antiderivative of $\ln(u)$ is $u\ln(u)-u$, which can be found through integration by parts, for example.  Therefore, this simplifies to
$$
\lim_{r\rightarrow 0^+}(u\ln u-u)|_1^r=\lim_{r\rightarrow 0^+}(r\ln r-r)-(1\ln 1-1).  
$$
Now, $\lim_{r\rightarrow 0^+}r\ln r$ is an indeterminate form, but an application of l'Hopital's rule results in a value of $0$, so this entire expression simplifies to $1$, as expected.
A: Note that
$$\ln(\frac{1}{1-x})=-\ln(1-x)$$
You can easily integrate this now, without a power series
A: Remember that the integral of the logarithm can be evaluated using integral by parts:
$$\int \ln (x) dx = \int \frac{dx}{dx}\ln (x) dx = x\ln(x) - \int x\frac{1}{x}dx = x\ln(x) -x + C$$
Now, using what Gabe said before, you can easily compute your integral.
