Question

The value of the integral $\int \frac{3x-1}{(x+2{)}^2}dx$ is
(where $C$ is an arbitrary constant)

• A. $\ln |x+2|+\frac{1}{(x+2)}+C$
• B. $3\ln |x+2|+\frac{1}{(x+2)}+C$
• C. $3\ln |x+2|+\frac{7}{(x+2)}+C$
• D. $\ln |x+2|+\frac{7}{(x-2)}+C$

Answer 1

The correct option is C $3\ln |x+2|+\frac{7}{(x+2)}+C$

Let $\frac{3x-1}{(x+2{)}^2}=\frac{A}{(x+2)}+\frac{B}{(x+2{)}^2}$
$\Rightarrow 3x-1=A(x+2)+B$
Equating the coefficient of $x$ and constant term, we obtain
$A=32A+B=-1\Rightarrow B=-7\therefore \frac{3x-1}{(x+2{)}^2}=\frac{3}{(x+2)}-\frac{7}{(x+2{)}^2}\Rightarrow \int \frac{3x-1}{(x+2{)}^2}dx=3\int \frac{1}{(x+2)}dx-7\int \frac{1}{(x+2{)}^2}dx=3\ln |x+2|+\frac{7}{(x+2)}+C$