Question

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

• A. $-\frac{{(x^4+1)}^{1/4}}{x}+c$
• B. $\frac{{(x^4+1)}^{1/4}}{x}+c$
• C. zero
• D. none of the above

Answer 1

Answer: (A)

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