JEE Mains · Maths · STD 12 - 7.1 indefinite integral
\(\smallint \frac{{dx}}{{{x^2}{{\left( {{x^4} + 1} \right)}^{\frac{3}{4}}}}} = \)
- A \( - {\left( {\frac{{{x^4} + 1}}{{{x^4}}}} \right)^{\frac{1}{4}}} + c\)
- B \({\left( {\frac{{{x^4} + 1}}{{{x^4}}}} \right)^{\frac{1}{4}}} + c\)
- C \({\left( {{x^4} + 1} \right)^{\frac{1}{4}}} + c\)
- D \( - {\left( {{x^4} + 1} \right)^{\frac{1}{4}}} + c\)
Answer & Solution
Correct Answer
(A) \( - {\left( {\frac{{{x^4} + 1}}{{{x^4}}}} \right)^{\frac{1}{4}}} + c\)
Step-by-step Solution
Detailed explanation
\(\int \frac{1 d x}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}\) Taking \(x^{4}\) common from denominator \(=\int \frac{1 d x}{x^{2}\left(x^{4}\right)^{\frac{3}{4}}\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}\)…
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