Question

Evaluate $\int \frac{x+3}{x^2+5x+4}dx$
(where $C$ is constant of integration)

• A. $\frac{1}{2}\ln |x^2+5x+4|-\frac{1}{6}\ln \left|\frac{x-1}{x-4}\right|+C$
• B. $-\frac{1}{2}\ln |x^2+5x+4|+\frac{1}{6}\ln \left|\frac{x+1}{x+4}\right|+C$
• C. $\frac{1}{2}\ln |x^2+5x+4|-\frac{1}{6}\ln \left|\frac{x+1}{x+4}\right|+C$
• D. $\frac{1}{2}\ln |x^2+5x+4|+\frac{1}{6}\ln \left|\frac{x+1}{x+4}\right|+C$

Answer 1

The correct option is D $\frac{1}{2}\ln |x^2+5x+4|+\frac{1}{6}\ln \left|\frac{x+1}{x+4}\right|+C$

Let $I=\int \frac{x+3}{x^2+5x+4}dx$
$\Rightarrow I=\frac{1}{2}\int \frac{2x+5}{x^2+5x+4}dx+(3-\frac{5}{2})\int \frac{dx}{(x+4)(x+1)}$

$\Rightarrow I=\frac{1}{2}\int \frac{2x+5dx}{x^2+5x+4x}+\frac{1}{2}\int \frac{dx}{(x+4)(x+1)}$

Let $I_1=\frac{1}{2}\int \frac{2x+5}{x^2+5x+4}dx$

Putting $x^2+5x+4=t$,

$(2x+5)dx=dt$

$\Rightarrow I_1=\frac{1}{2}\int \frac{dt}{t}=\frac{1}{2}\ln |x^2+5x+4|+C_1$

Now, $I_2=\frac{1}{2}\int \frac{dx}{(x+4)(x+1)}$

$\Rightarrow I_2=\frac{1}{2}\cdot \frac{1}{3}\int \frac{(x+4)-(x+1)}{(x+4)(x+1)}dx$

$=\frac{1}{6}\int \frac{dx}{x+1}dx-\frac{1}{6}\int \frac{dx}{x+4}$

$\Rightarrow I_2=\frac{1}{6}\ln \left|\frac{x+1}{x+4}\right|+C_2$

$\therefore I=\frac{1}{2}\ln |x^2+5x+4|+\frac{1}{6}\ln \left|\frac{x+1}{x+4}\right|+C$