Question

Evaluate $\int \left(\right(3x+2\left)\sqrt{x^2+3x+2}\right)dx$
(where $C$ is constant of integration)

• A. ${(x^2+3x+2)}^{1/2}+\frac{3(2x+3)}{16}\sqrt{x^2+3x+2}-\frac{5}{32}\ln |(x+\frac{3}{2})+\sqrt{x^2+3x+2}|+C$
• B. ${(x^2+3x+2)}^{3/2}+\frac{3(2x+3)}{16}\sqrt{x^2+3x+2}-\frac{3}{32}\ln |(x+\frac{3}{2})+\sqrt{x^2+3x+2}|+C$
• C. ${(x^2+3x+2)}^{3/2}-\frac{5(2x+3)}{8}\sqrt{x^2+3x+2}-\frac{5}{16}\ln |(x+\frac{3}{2})+\sqrt{x^2+3x+2}|+C$
• D. ${(x^2+3x+2)}^{3/2}+\frac{(2x+3)}{8}\sqrt{x^2+3x+2}-\frac{1}{16}\ln |(x+\frac{3}{2})+\sqrt{x^2+3x+2}|+C$

Answer 1

The correct option is C ${(x^2+3x+2)}^{3/2}-\frac{5(2x+3)}{8}\sqrt{x^2+3x+2}-\frac{5}{16}\ln |(x+\frac{3}{2})+\sqrt{x^2+3x+2}|+C$

Let $I=\int \left(\right(3x+2\left)\sqrt{x^2+3x+2}\right)dx$
$\Rightarrow I=\frac{3}{2}\int (2x+3)\sqrt{x^2+3x+2}dx+\int (2-\frac{9}{2})\sqrt{x^2+3x+2}dx$
Put, $x^2+3x+2=t$ in first integral, we get
$\Rightarrow I=\frac{3}{2}\int \sqrt{t}dt-\frac{5}{2}\int \sqrt{{(x+\frac{3}{2})}^2+(2-\frac{9}{4})}dx$
$\Rightarrow I=\frac{3}{2}\int \sqrt{t}dt-\frac{5}{2}\int \sqrt{{(x+\frac{3}{2})}^2-{\left(\frac{1}{2}\right)}^2}dx$
$\Rightarrow I=\frac{3}{2}\frac{t^{3/2}}{3/2}-\frac{5}{2}[\frac{(x+\frac{3}{2})}{2}\sqrt{x^2+3x+2}+\frac{1}{4}\cdot \frac{1}{2}\ln |(x+\frac{3}{2})+\sqrt{x^2+3x+2}|]+C\therefore I={(x^2+3x+2)}^{3/2}-\frac{5(2x+3)}{8}\sqrt{x^2+3x+2}-\frac{5}{16}\ln |(x+\frac{3}{2})+\sqrt{x^2+3x+2}|+C$