Find ∫ (3x+5)/(x^2+3x-18) dx

The correct answer is Option (1) → $\frac{13}{9} \ln|x+6| + \frac{14}{9} \ln|x-3| + C$

Let $I = \int \frac{3x+5}{x^2+3x-18} dx$

$I = \int \frac{3x+5}{(x+6)(x-3)} dx$

Let $\frac{3x+5}{(x+6)(x-3)} = \frac{A}{x+6} + \frac{B}{x-3}$

So, $3x+5 = A(x-3) + B(x+6)$

On comparing:

$A + B = 3$   …(i)

$-3A + 6B = 5$   …(ii)

$-3A+6(3-A)=5$

$-3A+18-6A=5$

$A =\frac{-13}{-9}= \frac{13}{9}$

and $B =3-A=3-\frac{13}{9}= \frac{14}{9}$

So, $\frac{3x+5}{(x+6)(x-3)} = \int \frac{13 dx}{9(x+6)} + \int \frac{14 dx}{9(x-3)}$

$= \frac{13}{9} \ln |x+6| + \frac{14}{9} \ln |x-3| + C$