Question

The value of $\int \frac{1}{x^2(x^4+1{)}^{3/4}}dx$ 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

Answer: (D)

$-{(1+\frac{1}{x^4})}^{1/4}+C$

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

$=\int \frac{1}{x^5(1+\frac{1}{x^4}{)}^{3/4}}dx$

Let $\frac{1}{x^4}+1=z$

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

$=\frac{-1}{4}\int \frac{dz}{z^{3/4}}=-\frac{1}{4}\frac{z^{-3/4}+1}{\frac{-3}{4}+1}+c$

$=-z^{\frac{1}{4}}+c$

$=-{(1+{\frac{1}{x}}^4)}^{1/4}+c$