Question

If $\int \frac{x^2+4}{x^4+16}dx=\frac{1}{k}{\tan }^{-1}\left(\frac{x^2-4}{kx}\right)+C$, then $k^2=$(where $k$ is fixed constant and $C$ is integration constant)

• A. 8.0
• B. 8
• C. 8.00
• D. 8.000

Answer 1

Answer: (A), (B), (C), (D)

$I=\int \frac{x^2+4}{x^4+16}dx$
Dividing $N^r$ and $D^r$ by $x^2$
$=\int \frac{1+\frac{4}{x^2}}{x^2+\frac{16}{x^2}}dx$
$=\int \frac{1+\frac{4}{x^2}}{x^2+{\left(\frac{4}{x}\right)}^2-8+8}$
$=\int \frac{1+\frac{4}{x^2}}{{(x-\frac{4}{x})}^2+8}dx$
Let $x-\frac{4}{x}=t$
$\Rightarrow (1+\frac{4}{x^2})dx=dt$
$\therefore I=\int \frac{dt}{t^2+(2\sqrt{2}{)}^2}=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{t}{2\sqrt{2}}\right)+C\{\because \int \frac{1}{t^2+a^2}dt=\frac{1}{a}{\tan }^{-1}\left(\frac{t}{a}\right)+C\}$
$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{x-\frac{4}{x}}{2\sqrt{2}}\right)+C$
$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-4}{\left(2\sqrt{2}\right)x}\right)+C$
$\Rightarrow k=2\sqrt{2}\Rightarrow k^2=8$