Evaluation of the integral $\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$ How can I evaluate the integral
$$\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$$ 
I tried manipulating the known integral
$$\int_0^1 \frac{\ln(1 - x)}{x}dx = -\frac{\pi^2}{6}$$
but couldn't do anything with it.  

You can use the integral you want to use, and the Dilogarithm function as mentioned in the comments.
Below we give a complete proof, including a derivation of the value of the integral you wanted to use.
The Dilogarithm function is defined as
$$\text{Li}_2(z) = -\int_{0}^{z} \frac{\log (1-x)}{x} \text{dx} = \sum_{n=1}^{\infty} \frac{z^n}{n^2}, \quad |z| \le 1$$ 
The integral which you want to use is $\displaystyle -\text{Li}_2(1)$.
Note that $\displaystyle \text{Li}_2(1) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \zeta(2) = \frac{\pi^2}{6}$. (For multiple proofs of that, see here: Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$)
In your integral(whose value you want), make the substitution $\displaystyle x = 2t -1$ and we get
$$\int_{\frac{1}{2}}^{1} \frac{\log (2(1-t))}{t} \text{dt} = \log^2 2 + \int_{\frac{1}{2}}^{1} \frac{\log (1-t)}{t} \text{dt} = \log^2 2 + \text{Li}_2 \left(\frac{1}{2} \right) - \text{Li}_2(1) $$
Now the Dilogarithm function also satisfies the identity
$$\text{Li}_2(x) + \text{Li}_2(1-x)  = \frac{\pi^2}{6}-\log x \log (1-x), 0 \lt x \lt 1$$
This identity can easily be proven by just differentiating and using the value of $\displaystyle \text{Li}_2(1)$: 
$$\text{Li}_2'(x) - \text{Li}_2'(1-x) = -\frac{\log (1-x)}{x} + \frac{\log x}{1-x} = (-\log x \log (1-x))'$$
and so 
$$\text{Li}_2(x) + \text{Li}_2(1-x)  = C -\log x \log (1-x), 0 \lt x \lt 1$$
Taking limits as $\displaystyle x \to 1$ gives us $\displaystyle C = \frac{\pi^2}{6}$.
Thus
$$\text{Li}_2(x) + \text{Li}_2(1-x)  = \frac{\pi^2}{6}-\log x \log (1-x), 0 \lt x \lt 1$$
Setting $\displaystyle x = \frac{1}{2}$ gives us the value of $\displaystyle \text{Li}_2\left(\frac{1}{2}\right) = \frac{\pi^2}{12} - \frac{\log^2 2}{2}$
Thus your integral is
$$\log^2 2 + \text{Li}_2 \left(\frac{1}{2} \right) - \text{Li}_2(1) = \frac{\log^2 2}{2} - \frac{\pi^2}{12}$$
