Q134 / 200 · Manipal (MET) 2010 · Mathematics
$\int \frac{x^2+4}{x^4+16} d x$ is equal to
- A$\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^2+4}{2 x}\right)+c$
- Correct answerB$\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^2-4}{2 \sqrt{2} x}\right)+c$
- C$\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^2+4}{2 \sqrt{2} x}\right)+c$
- D$\frac{1}{2} \tan ^{-1}\left(\frac{x^2-4}{2 x}\right)+c$
Solution
$\int \frac{x^2+4}{x^4+16} d x$
$=\int \frac{1+\frac{4}{x^2}}{x^2+\frac{16}{x^2}} d x=\int \frac{d\left(x-\frac{4}{x}\right)}{\left(x-\frac{4}{x}\right)^2+8}$
$=\int \frac{d t}{t^2+(2 \sqrt{2})^2}$, where $t=x-\frac{4}{x}$
$=\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^2-4}{2 \sqrt{2} x}\right)+c$
Hence, the correct answer is option: 'B'.