Find: ∫ (2x+3)/(x^2(x+3)) dx.

The correct answer is Option (2) → $\frac{1}{3} \ln \left| \frac{x}{x+3} \right| - \frac{1}{x} + C$

$I = \int \frac{2x+3}{x^2(x+3)} dx$ ...(i)

$\frac{2x+3}{x^2(x+3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+3}$ ...(ii)

$\frac{2x+3}{x^2(x+3)} =\frac{Ax(x+3) + B(x+3) + C(x^2)}{x^2(x+3)}$

$2x+3 = Ax(x+3) + B(x+3) + C(x^2)$

Put $x=0$:

$2(0)+3 = A(0) + B(3) + C(0)$

$3 = 0+3B+0$

$⇒B = 1$

Put $x=-3$

$2(-3)+3 = A(0) + B(0) + C(-3)^2$

$-3 =0+0+ 9C$

$⇒C = -\frac{1}{3}$

Put $x=1$:

$2(1)+3 = A(1)(4) + B(4) + C(1)$

$5=4A+4B+C$

$5 = 4A + 4(1) + (-\frac{1}{3})$

$4A = 5 - 4 + \frac{1}{3}$

$4A= 1 + \frac{1}{3}$

$4A= \frac{4}{3} ⇒A = \frac{1}{3}$

$∴\frac{2x+3}{x^2(x+3)} = \left( \frac{1/3}{x} + \frac{1}{x^2} - \frac{1/3}{x+3} \right)$

Now, $I = \int \frac{2x+3}{x^2(x+3)} dx$

$ = \int \left( \frac{1}{3x} + \frac{1}{x^2} - \frac{1}{3(x+3)} \right) dx$

$=\frac{1}{3}\int\frac{1}{x}dx+\int\frac{1}{x^2}dx-\frac{1}{3}\int\frac{1}{x+3}dx$

$I = \frac{1}{3} \ln|x| - \frac{1}{x} - \frac{1}{3} \ln|x+3| + C$