Question

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

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

Answer 1

Answer: (D)

$I=\int \frac{1}{x^2{\left(x^4+1\right)}^{3/4}}dx$
$=\int \frac{1}{x^2{\left[x^4\left(1+\frac{1}{x^4}\right)\right]}^{3/4}}dx$
$=\int \frac{1}{x^2\cdot x^3{\left(1+\frac{1}{x^4}\right)}^{3/4}}dx$
$I=\int \frac{1}{x^5{\left(1+\frac{1}{x^4}\right)}^{3/4}}dx$
Put $1+\frac{1}{x^4}=t$
$\frac{-4}{x^5}dx=dt$
$I=\int \frac{-dt}{4t^{3/4}}$
$=-\frac{1}{4}\int t^{-3/4}dt$
$I=-\frac{1}{4}\left[\frac{t^{-\frac{3}{4}+1}}{\frac{-3}{4}+1}\right]+C$
$I=\frac{-1}{4}\frac{t^{1/4}}{1/4}+C$
$I=-{\left(1+\frac{1}{{\lambda }^4}\right)}^{1/4}+C$