Question

Evaluate $\int \frac{9x+27}{x^3+3x^2+16x+48}dx$
(where $C$ is constant of integration)

• A. $\frac{9}{4}{\tan }^{-1}\frac{x}{4}+C$
• B. $\frac{9}{2}{\tan }^{-1}\frac{x}{4}+C$
• C. $\frac{3}{4}{\tan }^{-1}\frac{x}{4}+C$
• D. $\frac{1}{4}{\tan }^{-1}\frac{x}{4}+C$

Answer 1

The correct option is A $\frac{9}{4}{\tan }^{-1}\frac{x}{4}+C$

Let $I=\int \frac{9x+27}{x^3+3x^2+16x+48}dx$
$\Rightarrow I=\int \frac{9(x+3)}{(x^2+16)(x+3)}dx=9\int \frac{dx}{x^2+4^2}$

$=\frac{9}{4}{\tan }^{-1}\frac{x}{4}+C$