Integrable?

Maclurin expansion of $ln(1+x) = x - \frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+...$ where $-1<x<1$

Replace $x$ by $x-1$ in above formula

$ln(x)=(x-1)- \frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4+...$

Also by limits of integration $0<x<1$ means $-1<x-1<0$ So we can use this above formula

Now $\frac{\ln x}{x-1}=1- \frac{1}{2}(x-1)+\frac{1}{3}(x-1)^2-\frac{1}{4}(x-1)^3+...$

$\displaystyle I = \int_{0}^{1}\dfrac{\ln(x)}{x-1} \, dx = \int_{0}^{1} (1- \frac{1}{2}(x-1)+\frac{1}{3}(x-1)^2-\frac{1}{4}(x-1)^3+... ) \, dx$

Using $\int_a^b \mathrm f(x)\,\mathrm{d}x = \int_a^b \mathrm f(a+b-x)\,\mathrm{d}x$ we get

$\displaystyle I = \int_{0}^{1} (1- \frac{1}{2}(1-x-1)+\frac{1}{3}(1-x-1)^2-\frac{1}{4}(1-x-1)^3+... ) \, dx$

$\displaystyle I=\int_{0}^{1} (1+ \frac{1}{2}x+\frac{1}{3}x^2+\frac{1}{4}x^3+... ) \, dx$

$\displaystyle I= 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....=\frac{\pi^{2}}{6}$

You can check this above sum here

$100I = 100\frac{\pi^{2}}{6}=\boxed{164.493}$

$\displaystyle I = - \int_{0}^{1} \dfrac{ln(x)}{1-x}dx$
Using the Maclaurin series for $\dfrac{1}{1-x}$
$\displaystyle \therefore I = -\int_{0}^{1} ln(x) \times \left( 1 + x² + x³ ... \right) dx = -\int_{0}^{1} ln(x)\sum_{k= 0}^{\infty}x^{k} dx$
Interchanging the order of integration and summation,
$\displaystyle \therefore I = -\sum_{k=0}^{\infty} \int_{0}^{1} x^{k}ln(x)dx$
Now, it can be proved that,
$\displaystyle \int_{0}^{1} x^{m}ln^{n}(x)dx = \dfrac{(-1)^n \times n!}{(m+1)^{n+1}}$
Let m = k, n = 1,
$\displaystyle \therefore I = - \sum_{k=0}^{\infty} -\dfrac{1}{(k+1)^{2}} = \sum_{n= 1}^{\infty} \dfrac{1}{n^{2}}$
$\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^{2}}$ is the solution to the Basel Problem, and its value is equal to $\dfrac{\pi^{2}}{6}$
$\displaystyle \therefore I = \dfrac{\pi^{2}}{6} = 1.64493$
$\displaystyle \therefore 100I = 100 \times 1.64493 = 164.493$

If I am not wrong,the question asks the value of the dilogarithm function at 0,which is a standard result.

amazing solution