Question

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

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

Answer 1

The correct option is C $=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-4}{2x\sqrt{2}}\right)+C$

$I=\int \frac{x^2+4}{x^4+16}dx$
$=\int \frac{1+\frac{4}{x^2}}{x^2+\frac{16}{x^2}}dx$
$=\int \frac{1+\frac{4}{x^2}}{x^2+{\left(\frac{4}{x}\right)}^2-8+8}dx$
$=\int \frac{1+\frac{4}{x^2}}{{(x-\frac{4}{x})}^2+8}dx$
Put $x-\frac{4}{x}=t.$

$\Rightarrow (1+\frac{4}{x^2})dx=dt$
$\therefore I=\int \frac{dt}{t^2+{\left(2\sqrt{2}\right)}^2}$
$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{t}{2\sqrt{2}}\right)+C$
$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-4}{2x\sqrt{2}}\right)+C$