Solving an equation involving an integral: $\int_0^1\frac{ax+b}{(x^2+3x+2)^2}\:dx=\frac52.$ Determine a pair of number $a$ and $b$ for which 

$$\int_0^1\frac{ax+b}{(x^2+3x+2)^2}\:dx=\frac52.$$

I tried putting $x$ as $1-x$ as the integral wouldn't change but could not move forward from there so can you please suggest me what should I do next.

Hint. One may observe that
$$
(x^2+3x+2)=(x+1)(x+2)
$$ leading to the following partial fraction decomposition
$$
\begin{align}
\frac{ax+b}{(x^2+3x+2)^2}&=\frac{ax+b}{(x+1)^2(x+2)^2}
\\\\&=\frac{-a+b}{(x+1)^2}+\frac{3 a-2 b}{x+1}+\frac{-2 a+b}{(x+2)^2}+\frac{-3 a+2 b}{x+2}.
\end{align}
$$ Integrating each term gives
$$
\int_0^1\frac{ax+b}{(x^2+3x+2)^2}\:dx= \frac{-5a+4b}6+(3a-2b)(2\ln 2-\ln 3)
$$
Can you take it from here?

Edit. Partial fraction decomposition.
$$
\begin{align}
\frac{ax+b}{(x+1)^2(x+2)^2}&=\frac{A_1}{(x+1)^2}+\frac{B_1}{x+1}+\frac{A_2}{(x+2)^2}+\frac{B_2}{x+2}.
\end{align}
$$ By multiplying throughout by $(x+1)^2$ One has
$$
\begin{align}
\frac{ax+b}{(x+2)^2}&=A_1+(x+1)B_1+(x+1)^2\left(\frac{A_2}{(x+2)^2}+\frac{B_2}{x+2}\right).
\end{align}
$$ Putting $x=-1$ gives $A_1=-a+b$. Similarly by multiplying throughout by $(x+2)^2$ putting $x=-2$ gives $A_2=-2a+b$. Then by multiplying throughout by $(x+1)$ letting $x \to \infty$ gives $B_1+B_2=0$, finally $x=0$ gives $B_1=3a-2b=-B_2$. We are done.
