JEE Mains · Maths · STD 12 - 7.1 indefinite integral
If \(\int {\frac{{\sqrt {1 - {x^2}} }}{{{x^4}}}} dx\, = \,A\,(x)\,{(\sqrt {1 - {x^2}} )^m}\, + \,C,\) for a suitable chosen integer \(m\) and a function \(A(x),\) where \(C\) is a constant of integration, then \((A(x))^m\) equals
- A \(\frac{{ - 1}}{{27\,{x^9}}}\)
- B \(\frac{{ - 1}}{{3\,{x^3}}}\)
- C \(\frac{{ 1}}{{27\,{x^6}}}\)
- D \(\frac{{ 1}}{{9\,{x^4}}}\)
Answer & Solution
Correct Answer
(A) \(\frac{{ - 1}}{{27\,{x^9}}}\)
Step-by-step Solution
Detailed explanation
\(\int \frac{\sqrt{1-x^{2}}}{x^{4}} d x=\int \frac{x \sqrt{\frac{1}{x^{2}}-1}}{x^{4}} d x=\int \frac{1}{x^{3}} \sqrt{\frac{1}{x^{2}}-1} d x\) \({\rm{ Let }}\frac{1}{{{x^2}}} - 1 = t\) \( - \frac{2}{{{x^3}}}{\rm{d}}x = {\rm{dt}}\)…
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