Question

Evaluate $\int \frac{x^{2}+4}{x^{4}+16}dx.$

• A. $\frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x^{2}-4}{2x\sqrt{2}}\right)+c$
• B. $\frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x^{2}-4}{2\sqrt{2}}\right)+c$
• C. $\frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x^{2}-4}{x\sqrt{2}}\right)+c$
• D. None of the above

Answer 1

Answer: (A)

Let $I=\int \frac{x^{2}+4}{x^{4}+16} d x$
$=\int \frac{1+\frac{4}{x^{2}}}{x^{2}+\frac{16}{x^{2}}}$
$d x=\int \frac{1+\frac{4}{x^{2}}}{\left(x-\frac{4}{x}\right)^{2}+8} d x$
Putting $x-\frac{4}{x}=t$
So that $\left(1+\frac{4}{x^{2}}\right) d x=d t$
$\therefore I=\int \frac{d t}{t^{2}+(2 \sqrt{2})^{2}}$
$\Rightarrow \frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{t}{2 \sqrt{2}}\right)+C$
$\Rightarrow I=\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{4}{x}}{2 \sqrt{2}}\right)+C$
$=\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^{2}-4}{2 x \sqrt{2}}\right)+C$