Question

The integral $\int \frac{dx}{x^2{\left(x^4+1\right)}^{3/4}}$ equals

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

Answer 1

Answer: (D)

$\int \frac{dx}{x^2{\left(x^4+1\right)}^{3/4}}$
$=\int \frac{dx}{x^5{\left(1+x^{-4}\right)}^{3/4}}=\int \frac{x^{-5}}{{\left(1+x^{-4}\right)}^{3/4}}dx\dots (1)$
put $1+x^{-4}=T^4$
$-4x^{-5}dx=4T^{3}dT$
$\Rightarrow$ (1) become
$-\int \frac{T^{3}dT}{T^3}=-T+C$
$=-{\left(1+x^{-4}\right)}^{1/4}+C$
$=-{\left(\frac{1+x^4}{x^4}\right)}^{1/4}+C$