Question

$\int (x+1)\sqrt{2x^2+3}dx$ is equal to
(where $C$ is integration constant)

• A. $\frac{1}{3}{(x^2+3)}^{\frac{3}{2}}+\frac{x}{2}\sqrt{2x^2+3}+\frac{3}{2\sqrt{2}}\ln |x+\sqrt{x^2+\frac{2}{3}}|+C$
• B. $\frac{1}{6}{(2x^2+3)}^{\frac{3}{2}}+\frac{x}{2}\sqrt{2x^2+3}+\frac{3}{2\sqrt{2}}\ln |x+\sqrt{x^2+\frac{3}{2}}|+C$
• C. $\frac{1}{6}{(x^2-3)}^{\frac{3}{4}}+\frac{x}{2}\sqrt{x^2-3}+\frac{3}{2\sqrt{2}}\ln |x+\sqrt{x^2+\frac{3}{2}}|+C$
• D. $\frac{1}{3}{(x^2-6)}^{\frac{3}{4}}+\frac{x}{2}\sqrt{x^2-6}+\frac{3}{2\sqrt{2}}\ln |x+\sqrt{x^2+\frac{3}{2}}|+C$

Answer 1

The correct option is B $\frac{1}{6}{(2x^2+3)}^{\frac{3}{2}}+\frac{x}{2}\sqrt{2x^2+3}+\frac{3}{2\sqrt{2}}\ln |x+\sqrt{x^2+\frac{3}{2}}|+C$

$I=\int (x+1)\sqrt{2x^2+3}dx$
Let $x+1=\lambda \frac{d}{dx}(2x^2+3)+\mu$
$\Rightarrow x+1=4x\lambda +\mu$
Equating coefficient of same degree term, we get
$\Rightarrow \lambda =\frac{1}{4},\mu =1\Rightarrow I=\int \left(\frac{1}{4}\right(4x)+1)\sqrt{2x^2+3}dx\Rightarrow I=\frac{1}{4}\int 4x\sqrt{2x^2+3}dx+\int \sqrt{2x^2+3}dx$

Putting $2x^2+3=t\Rightarrow 4xdx=dt$
Now,
$\Rightarrow I=\frac{1}{4}\int \sqrt{t}dt+\int \sqrt{2x^2+3}dx=\frac{t^{\frac{3}{2}}}{6}+\int \sqrt{2[x^2+{\left(\sqrt{\frac{3}{2}}\right)}^2]}dx\Rightarrow I=\frac{1}{6}{(2x^2+3)}^{\frac{3}{2}}+\sqrt{2}\int \sqrt{x^2+{\left(\sqrt{\frac{3}{2}}\right)}^2}dx\therefore I=\frac{1}{6}{(2x^2+3)}^{\frac{3}{2}}+\frac{x}{2}\sqrt{2x^2+3}+\frac{3}{2\sqrt{2}}\ln |x+\sqrt{x^2+\frac{3}{2}}|+C$