Question

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

• A. ${(1+\frac{1}{x^4})}^{1/4}+C$
• B. ${(x^4+1)}^{1/4}+C$
• C. ${(1-\frac{1}{x^4})}^{1/4}+C$
• D. $-{(1+\frac{1}{x^4})}^{1/4}+C$

Answer 1

The correct option is D $-{(1+\frac{1}{x^4})}^{1/4}+C$

$I=\int \frac{1}{x^2{(x^4+1)}^{\frac{3}{4}}}dx$
$=\int \frac{1}{x^2{\left[x^4(1+\frac{1}{x^4})\right]}^{\frac{3}{4}}}dx$
$=\int \frac{1}{x^2.x^3{(1+\frac{1}{x^4})}^{\frac{3}{4}}}dx$
$=\int x^{-5}{(1+x^{-4})}^{\frac{-3}{4}}dx$
Let $y=1+x^{-4}$
$\frac{dy}{dx}=-4x^{-5}$

$x^{-5}dx=\frac{-dy}{4}$

$\therefore I=\int -\frac{1}{4}{\left(y\right)}^{\frac{-3}{4}}dy$

$=-\frac{1}{4}\frac{y^{\frac{-3}{4}+1}}{\frac{-3}{4}+1}+C$

$=-\frac{1}{4}\frac{y^{\frac{1}{4}}}{\left(\frac{1}{4}\right)}+C$

$=-y^{\frac{1}{4}}+C$
$=-{[1+\frac{1}{x^4}]}^{\frac{1}{4}}+C$