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Q.
If $\int \frac{1}{6 x^2+13 x+6} d x=\frac{1}{a} \ln \left|\frac{b x+2}{2 x+b}\right|+C$, where $a, b \in R$ and $C$ is constant of integration to is equal to
Integrals
Solution:
Given integral $=\int \frac{1}{(2 x+3)(3 x+2)} d x=\int\left(\frac{\frac{-2}{5}}{(2 x+3)}+\frac{\frac{3}{5}}{(3 x+2)}\right) d x $
$I=\frac{-1}{5} \ln (2 x+3)+\frac{1}{5} \ln (3 x+2)+C $
$=\frac{1}{5} \ln \left|\frac{3 x +2}{2 x +3}\right|+ C $
$\therefore a =5 \text { and } b =3 \Rightarrow a + b =8 $