# Evaluate the following integral:

Question:

Evaluate the following integral:

$\int \frac{2 x+1}{(x-2)(x-3)} d x$

Solution:

Let, $I=\int \frac{2 x+1}{(x-2)(x-3)} d x$

Now, let $\frac{2 x+1}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}$

$\Rightarrow 2 x+1=A(x-3)+B(x-2)$

$\Rightarrow 2 x+1=(A+B) x-3 A-2 B$

Equating similar terms, we get,

$A+B=2$ and $3 A+2 B=-1$

So, $A=-5, B=7$

$\therefore \mathrm{I}=-5 \int \frac{\mathrm{dx}}{\mathrm{x}-2}+7 \int \frac{\mathrm{dx}}{\mathrm{x}-3}$

$\Rightarrow 1=-5 \log |x-2|+7 \log |x-3|+c$

$\Rightarrow 1=\log |x-2|^{-5}+\log |x-3|^{7}+c$

$\Rightarrow I=\log \left|\frac{(x-3)^{7}}{(x-2)^{5}}\right|+c$

Hence, $\int \frac{2 x+1}{(x-2)(x-3)} d x=\log \left|\frac{(x-3)^{7}}{(x-2)^{5}}\right|+c$