Question

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

• A. $\frac{1}{3}\sqrt{3x^2+2x+5}-\frac{4}{\sqrt{3}}\ln |(x+\frac{1}{3})+\sqrt{x^2-\frac{3x}{2}+\frac{5}{3}}|+C$
• B. $\frac{1}{3}\sqrt{3x^2+2x+5}-\frac{7}{\sqrt{3}}\ln |(x+\frac{1}{3})+\sqrt{x^2-\frac{3x}{2}+\frac{5}{3}}|+C$
• C. $\frac{2}{3}\sqrt{3x^2+2x+5}-\frac{7}{3\sqrt{3}}\ln |(x+\frac{1}{3})+\sqrt{x^2-\frac{2x}{3}+\frac{5}{3}}|+C$
• D. $\frac{2}{3}\sqrt{3x^2+2x+5}+\frac{7}{3\sqrt{3}}\ln |(x+\frac{1}{3})+\sqrt{x^2+\frac{2x}{3}+\frac{5}{3}}|+C$

Answer 1

The correct option is D $\frac{2}{3}\sqrt{3x^2+2x+5}+\frac{7}{3\sqrt{3}}\ln |(x+\frac{1}{3})+\sqrt{x^2+\frac{2x}{3}+\frac{5}{3}}|+C$

Let $I=\int \frac{2x+3}{\sqrt{3x^2+2x+5}}dx$
$I=\frac{1}{3}\int \frac{6x+2}{\sqrt{3x^2+2x+5}}dx+(3-\frac{2}{3})\int \frac{dx}{\sqrt{3x^2+2x+5}}$

$I=\frac{1}{3}\cdot 2\sqrt{3x^2+2x+5}+\frac{7}{3}\times \frac{1}{\sqrt{3}}\int \frac{dx}{\sqrt{x^2+\frac{2}{3}x+\frac{5}{3}}}$

$\Rightarrow I=\frac{2}{3}\sqrt{3x^2+2x+5}+\frac{7}{3\sqrt{3}}\int \frac{dx}{\sqrt{{(x+\frac{1}{3})}^2+{\left(\sqrt{\frac{14}{9}}\right)}^2}}$

$\therefore I=\frac{2}{3}\sqrt{3x^2+2x+5}+\frac{7}{3\sqrt{3}}\ln |(x+\frac{1}{3})+\sqrt{x^2+\frac{2x}{3}+\frac{5}{3}}|+C$