MCQ
$\int \frac{d x}{x^3\left(1+x^4\right)^{\frac{1}{2}}} \quad$ equals
  • A
    $-\frac{1}{2 x^2} \sqrt{1+x^4}+c$
  • $\frac{1}{2 x} \sqrt{1+x^4}+c$
  • C
    $-\frac{1}{4 x} \sqrt{1+x^4}+c$
  • D
    $\frac{1}{4 x^2} \sqrt{1+x^4}+c$

Answer

Correct option: B.
$\frac{1}{2 x} \sqrt{1+x^4}+c$
$\int \frac{d x}{x^3\left(1+x^4\right)^{\frac{1}{2}}}=\int \frac{d x}{x^5\left(1+\frac{1}{x^4}\right)^{\frac{1}{2}}}$
$\left(\text { Let } 1+x^{-4}=1+\frac{1}{x^4}=t, d t=-4 x^{-5} d x=-\frac{4}{x^5} d x \Rightarrow \frac{d x}{x^5}=-\frac{1}{4} d t \text { ) }\right.$
$=-\frac{1}{4} \int \frac{d t}{t^{\frac{1}{2}}}=-\frac{1}{4} \times 2 \times \sqrt{t}+c,$
 where $'c\ '$ denotes any arbitrary constant of integration. 
$=-\frac{1}{2} \sqrt{1+\frac{1}{x^4}}+c$

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