JEE Exam > JEE Questions > ∫tan-1[(2x)/(1-x2)]dx=a)x tan-1x+cb)x tan...
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∫tan-1[(2x)/(1-x2)]dx=
- a)x tan-1x+c
- b)x tan-1x-log(1+x2)+c
- c)2xtan-1x+log(1+x2)+c
- d)2x tan-1x-log(1+x2)+c
Correct answer is option 'D'. Can you explain this answer?
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∫tan-1[(2x)/(1-x2)]dx=a)x tan-1x+cb)x tan-1x-log(1+x2)+cc)2xtan-1x...
Integration by substitution:
- To solve the given integral, let's start by making a substitution. Let \(u = \tan^{-1}(x)\), which implies \(x = \tan(u)\) and \(dx = \sec^2(u) du\).
Applying the substitution:
- Substitute \(x = \tan(u)\) and \(dx = \sec^2(u) du\) into the integral:
\[
\int \tan^{-1}\left(\frac{2x}{1-x^2}\right)dx = \int \tan^{-1}\left(\frac{2\tan(u)}{1-\tan^2(u)}\right)\sec^2(u)du
\]
- Simplify the expression:
\[
= \int \tan^{-1}(\tan(2u))\sec^2(u)du
= \int 2u\sec^2(u)du
\]
Integrate the simplified expression:
- Integrate \(2u\sec^2(u)du\) with respect to \(u\) to get:
\[
= 2\tan(u) + C
= 2\tan(\tan^{-1}(x)) + C
= 2x + C
\]
- Therefore, the correct answer is \(2x \tan^{-1}(x) + C\), which corresponds to option D.
- To solve the given integral, let's start by making a substitution. Let \(u = \tan^{-1}(x)\), which implies \(x = \tan(u)\) and \(dx = \sec^2(u) du\).
Applying the substitution:
- Substitute \(x = \tan(u)\) and \(dx = \sec^2(u) du\) into the integral:
\[
\int \tan^{-1}\left(\frac{2x}{1-x^2}\right)dx = \int \tan^{-1}\left(\frac{2\tan(u)}{1-\tan^2(u)}\right)\sec^2(u)du
\]
- Simplify the expression:
\[
= \int \tan^{-1}(\tan(2u))\sec^2(u)du
= \int 2u\sec^2(u)du
\]
Integrate the simplified expression:
- Integrate \(2u\sec^2(u)du\) with respect to \(u\) to get:
\[
= 2\tan(u) + C
= 2\tan(\tan^{-1}(x)) + C
= 2x + C
\]
- Therefore, the correct answer is \(2x \tan^{-1}(x) + C\), which corresponds to option D.
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∫tan-1[(2x)/(1-x2)]dx=a)x tan-1x+cb)x tan-1x-log(1+x2)+cc)2xtan-1x+log(1+x2)+cd)2x tan-1x-log(1+x2)+cCorrect answer is option 'D'. Can you explain this answer?
Question Description
∫tan-1[(2x)/(1-x2)]dx=a)x tan-1x+cb)x tan-1x-log(1+x2)+cc)2xtan-1x+log(1+x2)+cd)2x tan-1x-log(1+x2)+cCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about ∫tan-1[(2x)/(1-x2)]dx=a)x tan-1x+cb)x tan-1x-log(1+x2)+cc)2xtan-1x+log(1+x2)+cd)2x tan-1x-log(1+x2)+cCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ∫tan-1[(2x)/(1-x2)]dx=a)x tan-1x+cb)x tan-1x-log(1+x2)+cc)2xtan-1x+log(1+x2)+cd)2x tan-1x-log(1+x2)+cCorrect answer is option 'D'. Can you explain this answer?.
∫tan-1[(2x)/(1-x2)]dx=a)x tan-1x+cb)x tan-1x-log(1+x2)+cc)2xtan-1x+log(1+x2)+cd)2x tan-1x-log(1+x2)+cCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about ∫tan-1[(2x)/(1-x2)]dx=a)x tan-1x+cb)x tan-1x-log(1+x2)+cc)2xtan-1x+log(1+x2)+cd)2x tan-1x-log(1+x2)+cCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ∫tan-1[(2x)/(1-x2)]dx=a)x tan-1x+cb)x tan-1x-log(1+x2)+cc)2xtan-1x+log(1+x2)+cd)2x tan-1x-log(1+x2)+cCorrect answer is option 'D'. Can you explain this answer?.
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