Question

$\int \frac{x^2}{(x^2+2)(x^2+3)}dx=$

• A. $-\sqrt{2}{\tan }^{-1}x+\sqrt{3}{\tan }^{1}x+c$
• B. $-\sqrt{2}{\tan }^{-1}\left(\frac{x}{\sqrt{2}}\right)+\sqrt{3}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)+c$
• C. $\sqrt{2}{\tan }^{-1}\left(\frac{x}{\sqrt{2}}\right)+\sqrt{3}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)+c$
• D. None of these

Answer 1

The correct option is B $-\sqrt{2}{\tan }^{-1}\left(\frac{x}{\sqrt{2}}\right)+\sqrt{3}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)+c$

$\int \frac{x^2}{(x^2+2)(x^2+3)}dx=\int x^2[\frac{1}{x^2+2}-\frac{1}{x^2+3}]dx$

$=\int (\frac{x^2}{x^2+2}-\frac{x^2}{x^2+3})dx$

$=\int (1-\frac{2}{x^2+2}-1+\frac{3}{x^2+3})dx$

$=\int (\frac{3}{x^2+3}-\frac{2}{x^2+2})dx$

$=3\left(\frac{1}{\sqrt{3}}\right)ta{n}^{-1}\frac{x}{\sqrt{3}}-2\left(\frac{1}{\sqrt{2}}\right)+ta{n}^{-1}\frac{x}{\sqrt{2}}$

$=\sqrt{3}ta{n}^{-1}\frac{x}{\sqrt{3}}-\sqrt{2}ta{n}^{-1}\frac{x}{\sqrt{2}}.$