Q112 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics

$\int \frac{d x}{\left(x^2+1\right)\left(x^2+4\right)}=$

  • A
    $\frac{1}{3} \tan ^{-1} x+\frac{1}{6} \tan ^{-1}\left(\frac{x}{2}\right)+c$
  • B
    $\frac{1}{3} \tan ^{-1} x-\frac{1}{3} \tan ^{-1}\left(\frac{x}{2}\right)+c$
  • C
    $\frac{1}{3} \tan ^{-1} x+\frac{1}{3} \tan ^{-1}\left(\frac{x}{2}\right)+c$
  • Correct answer
    D
    $\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1}\left(\frac{x}{2}\right)+c$

Solution

$\int \frac{d x}{\left(x^2+1\right)\left(x^2+4\right)}$ $\begin{aligned} & =\frac{1}{3} \int\left(\frac{1}{x^2+1}-\frac{1}{x^2+4}\right) d x \\ & =\frac{1}{3}\left[\int \frac{d x}{1+x^2}-\int \frac{d x}{4+x^2}\right] \\ & =\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1}\left(\frac{x}{2}\right)+C\end{aligned}$

Hence, the correct answer is option: 'D'.

Relevant Questions