Question

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

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