JEE Mains · Maths · STD 12 - 7.1 indefinite integral
The integral \(\int {x\,{{\cos }^{ - 1}}\,\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)dx} \,\left( {x > 0} \right)\) is equal to
- A \(- x + ( 1 + x^2)\, tan^{-1} \,x + c\)
- B \(x - (1 + x^2) cot^{-1} \,x + c\)
- C \(- x + ( 1 + x^2 ) cot^{-1} \,x + c\)
- D \(x - (1 + x^2) tan^{-1} \,x + c\)
Answer & Solution
Correct Answer
(A) \(- x + ( 1 + x^2)\, tan^{-1} \,x + c\)
Step-by-step Solution
Detailed explanation
(a) Let \(I=\int x \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) d x\) \(\therefore {\rm{I}} = 2\int {\mathop x\limits_{II} } \mathop {{{\tan }^{ - 1}}}\limits_I xdx\) Applying Integration by parts…
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