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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The value of ∫ dx x( x^ n + 1 ) is equal to

The value of ∫dxx(xn+1)∫ \dfrac{dx}{x( x^{n} + 1 )} is equal to

Options

  1. A1nlog⁡e(xnxn+1)+c\dfrac{1}{n} \log_{\mathrm{e}} ( \dfrac{x^{n}}{x^{n} + 1} ) + c
  2. B1nlog⁡e(xn+1xn)+c\dfrac{1}{n} \log_{\mathrm{e}} ( \dfrac{x^{n} + 1}{x^{n}} ) + c
  3. Clog⁡e(xnxn+1)+c\log_{\mathrm{e}} ( \dfrac{x^{n}}{x^{n} + 1} ) + c
  4. DNone of these

Correct answer

A. 1nlog⁡e(xnxn+1)+c\dfrac{1}{n} \log_{\mathrm{e}} ( \dfrac{x^{n}}{x^{n} + 1} ) + c

Step-by-step solution

I=∫dx(1+1xn)×xn+1\text{I=} ∫ \dfrac{dx}{( 1 + \dfrac{1}{x^{n}} ) \times x^{n + 1}}

put 1+1xn=t1 + \dfrac{1}{x^{n}} = t

−nxn+1.dx=dt- \dfrac{n}{x^{n + 1}}. dx = dt

I=∫−dtnt=−1nln∣t∣+cI = ∫ \dfrac{- \dfrac{dt}{n}}{t} = - \dfrac{1}{n} ln| t | + c

I=−1nln⁡(1+1xn)+cI = - \dfrac{1}{n} \ln( 1 + \dfrac{1}{x^{n}} ) + c

I=1nln⁡(xnxn+1)+cI = \dfrac{1}{n} \ln( \dfrac{x^{n}}{x^{n} + 1} ) + c
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