Question

$\int \frac{1+x^4}{(1-x^4{)}^{3/2}}dx$ is equal to

• A. $\frac{1}{\sqrt{x^2-\frac{1}{x^2}}}+C$
• B. $\frac{1}{\sqrt{\frac{1}{x^2}-x^2}}+C$
• C. $\frac{1}{\sqrt{\frac{1}{x^2}+x^2}}+C$
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{\sqrt{\frac{1}{x^2}-x^2}}+C$

Let $I=\int \frac{1+x^4}{(1-x^4{)}^{3/2}}dx=\int \frac{x^3(x+\frac{1}{x^3})dx}{(1-x^4{)}^{3/2}}$
$=\int \frac{(x+\frac{1}{x^3})dx}{{(\frac{1}{x^2}-x^2)}^{3/2}}$
Let, $\frac{1}{x^2}-x^2=t$
$\Rightarrow (\frac{-2}{x^3}-2x)dx=dt$
$\Rightarrow x+\frac{1}{x^3}dx=\frac{-1}{2}dt$
$\Rightarrow I=-\frac{1}{2}\int \frac{dt}{t^{3/2}}=\frac{1}{\sqrt{t}}+C=\frac{1}{\sqrt{\frac{1}{x^2}-x^2}}+C$.