Question

Evaluate: $\int \sqrt{x^2+3x}dx$

• A. $(\frac{x}{2}+\frac{3}{4})\sqrt{x^2+3x}+\frac{9}{8}{\cosh }^{-1}(\frac{2x}{3}+1)+c$
• B. $(\frac{x}{2}+\frac{3}{4})\sqrt{x^2+3x}-\frac{9}{8}{\cosh }^{-1}(\frac{2x}{3}+1)+c$
• C. $(\frac{x}{2}-\frac{3}{4})\sqrt{x^2+3x}-\frac{9}{8}{\cosh }^{-1}(\frac{2x}{3}-1)+c$
• D. $(\frac{x}{2}-\frac{3}{4})\sqrt{x^2-3x}+\frac{9}{8}{\cosh }^{-1}(\frac{2x}{3}-1)+c$

Answer 1

The correct option is B $(\frac{x}{2}+\frac{3}{4})\sqrt{x^2+3x}-\frac{9}{8}{\cosh }^{-1}(\frac{2x}{3}+1)+c$

Here $x^2+3x=x^2+2\times x\times \frac{3}{2}+{\left(\frac{3}{2}\right)}^2-{\left(\frac{3}{2}\right)}^2={(x+\frac{3}{2})}^2-{\left(\frac{3}{2}\right)}^2$

$\therefore I=\int \sqrt{{(x+\frac{3}{2})}^2-{\left(\frac{3}{2}\right)}^2}dx$ .... $\because \int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}{\cosh }^{-1}\left(\frac{x}{a}\right)+c$

Thus, $I=\frac{(x+\frac{3}{2})}{2}\sqrt{{\left(x+\frac{3}{2}\right)}^2-{\left(\frac{3}{2}\right)}^2}-\frac{{\left(\frac{3}{2}\right)}^2}{2}{\cosh }^{-1}\left(\frac{x+\frac{3}{2}}{\frac{3}{2}}\right)+c$

$\Rightarrow I=(\frac{x}{2}+\frac{3}{4})\sqrt{x^2+3x}-\frac{9}{8}{\cosh }^{-1}(\frac{2x}{3}+1)+c$