Question

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

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

Answer 1

Answer: (D)

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

Answer 2

$\int \limits \frac {dx}{x^2(x^4+1)^{3/4}}= \int \limits \frac {dx}{x^5 \bigg (1+ \frac {1}{x^4} \bigg )^{3/4}}$
put $1+ \frac {1}{x^4}=t^4 \Rightarrow \frac {-4}{x^5}dx=4t^3dt \Rightarrow \frac {dx}{x^5}=-t^3dt$
Hence, the integral becomes
$\int \limits \frac {-t^3dt}{t^3}=- \int \limits dt=-t+c=- \bigg (1+ \frac {1}{x^4} \bigg )^{1/4}+ c$