Question

Evaluate $\int \frac{(4x+1)dx}{x^2+3x+2}$.

• A. $2\log |x^2+3x+2|-5\log \left|\frac{x+1}{x+2}\right|+c$
• B. $-2\log |x^2+3x+2|-5\log \left|\frac{x+1}{x+2}\right|+c$
• C. $2\log |x^2+3x+2|+5\log \left|\frac{x+1}{x+2}\right|+c$
• D. $-2\log |x^2+3x+2|+5\log \left|\frac{x+1}{x+2}\right|+c$

Answer 1

The correct option is A $2\log |x^2+3x+2|-5\log \left|\frac{x+1}{x+2}\right|+c$

From the previous question we know,$\frac{4x+1}{x^2+3x+2}=\frac{2(2x+3)-5}{x^2+3x+2}$
Now, we can express the integral as
$I=\int \frac{2(2x+3)-5dx}{x^2+3x+2}$
$=2\int \frac{(2x+3)dx}{x^2+3x+2}-5\int \frac{dx}{x^2+3x+2}$
$=2\log \left(x^2+3x+2\right|-5\int \frac{dx}{x^2+3x+(9/4)-(9/4)+2}+c$
$=2\log \left(x^2+3x+2\right|-5\int \frac{dx}{(x+3/2{)}^2-(1/2{)}^2}+c$
$=2\log \left(x^2+3x+2\right|-5\frac{1}{2(1/2)}\log \left(\frac{x+\frac{3}{2}-\frac{1}{2}}{x+\frac{3}{2}+\frac{1}{2}}\right|+c$
$=2\log \left(x^2+3x+2\right|-5\log \left(\frac{x+1}{x+2}\right|+c$