Question
Evaluate $\int_{1}^{2} \frac{x d x}{(x+1)(x+2)}$ 

Answer

Let $I=\int_{1}^{2} \frac{x d x}{(x+1)(x+2)}$ 
Using partial fraction, we get $\frac{x}{(x+1)(x+2)}=\frac{-1}{x+1}+\frac{2}{x+2}$ 
So, $\int \frac{x d x}{(x+1)(x+2)}$ = - log |x + 1| +  2log |x + 2| = F(x)
Therefore, by the second fundamental theorem of calculus, we have
I = F(2) - F(1) = [- log 3 + 2 log 4] - [- log 2 + 2 log 3]
= -3 log 3 + log 2 + 2 log 4 = log$ \left(\frac{32}{27}\right)$ 

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