Question

If $I={\int }_0^1\frac{dx}{x^2{(1+x^4)}^{3/4}}$ is equal to

• A. $-\frac{{(1+x^4)}^{1/4}}{x}+c$
• B. $\frac{{(1+x^4)}^{1/4}}{x}+c$
• C. $-\frac{{(1+x^4)}^{3/4}}{x}+c$
• D. None of these

Answer 1

Answer: (A)

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

$I={\int }_0^1\frac{dx}{x^2{(1+x^4)}^{3/4}}=\int \frac{dx}{x^5{(1+\frac{1}{x^4})}^{3/4}}$
Put $1+\frac{1}{x^4}=t\Rightarrow \frac{dx}{x^5}=-\frac{1}{4}dt$
$\therefore I=-\frac{1}{4}\int \frac{dt}{t^{3/4}}=-\frac{1}{4}.\frac{t^{1/4}}{1/4}+c$
$=-{(1+\frac{1}{x^4})}^{1/4}+c=-\frac{{(1+x^4)}^{1/4}}{x}+c$