JEE Mains · Maths · STD 12 - 7.1 indefinite integral
If \(\int {\frac{{dx}}{{{x^3}{{\left( {1 + {x^6}} \right)}^{2/3}}}} = xf\left( x \right){{\left( {1 + {x^6}} \right)}^{\frac{1}{3}}} + C} \) where \(C\) is a constant of integration, then the function \(f(x)\) is equal to
- A \( - \frac{1}{{2{x^2}}}\)
- B \( - \frac{1}{{2{x^3}}}\)
- C \( + \frac{1}{{2{x^3}}}\)
- D \( \frac{3}{{{x^2}}}\)
Answer & Solution
Correct Answer
(B) \( - \frac{1}{{2{x^3}}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{I}=\int \frac{\mathrm{dx}}{\mathrm{x}^{3}\left(1+\mathrm{x}^{6}\right)^{2 / 3}}\) \( = \int {\frac{{dx}}{{{x^7}{{\left( {1 + \frac{1}{{{x^6}}}} \right)}^{2/3}}}}} \) \(\text { Put } 1+x^{-6}=t \) \(\Rightarrow \frac{d x}{x^{7}}=\frac{-d t}{6}\)…
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