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)

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