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