If the value of Definite Integral $\displaystyle \int_{0}^{2}\frac{ax+b}{(x^2+5x+6)^2}dx = \frac{7}{30}\;,$ Then value of $a^2+b^2$
$\bf{My\; Try::}$ We can write it as $$\frac{1}{2}\int_{0}^{2}\frac{a(2x+5)+(2b-5a)}{(x^2+5x+6)^2}dx = \frac{7}{30}$$
So we get $$\int_{0}^{2}\frac{a(x^2+5x+6)^{'}}{(x^2+5x+6)^2}dx+{(2b-5a)}\int_{0}^{2}\frac{1}{(x+2)^2(x+3)^2}dx = \frac{7}{15}$$
So we get $$-a\left[\frac{1}{(x^2+5x+6)}\right]_{0}^{2}+(2b-5a)\int_{0}^{2}......=\frac{7}{15}$$
Now How can I solve second Integral, Help Required, Thanks