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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

∫ (x + 1) - ln x x (x + 1) dx is equal to (where C is an arbitarary...

∫ \dfrac{\ln \left(x + 1\right) -\text{ ln } x}{x\left(x + 1\right)}dx is equal to (where CC is an arbitarary constant)

Options

  1. A−12[ln⁡(x+1x)]2+C- \dfrac{1}{2}{\left[\ln \left(\dfrac{x + 1}{x}\right)\right]}^{2} + C
  2. BC - \left[{ \left\{\ln \left(x + 1\right)\right\}}^{2} - (lnx){\left(ln x\right)}^{2}\right]$$
  3. C−ln⁡[ln⁡(x+1x)]+C- \ln \left[\ln \left(\dfrac{x + 1}{x}\right)\right] + C
  4. D−ln⁡(x+1x)+C- \ln \left(\dfrac{x + 1}{x}\right) + C

Correct answer

A. −12[ln⁡(x+1x)]2+C- \dfrac{1}{2}{\left[\ln \left(\dfrac{x + 1}{x}\right)\right]}^{2} + C

Step-by-step solution

Put ln⁡(x+1)−lnx=t\ln \left(x + 1\right) - ln x = t

⇒1x+1−1x=dtdx⇒x−(x+1)x(x+1)=dtdx⇒ \dfrac{1}{x + 1} - \dfrac{1}{x} = \dfrac{dt}{dx} ⇒ \dfrac{x - \left(x + 1\right)}{x\left(x + 1\right)} = \dfrac{dt}{dx}

⇒−dxx(x+1)=dt⇒dxx(x+1)=−dt⇒ \dfrac{- dx}{x\left(x + 1\right)} = dt ⇒ \dfrac{dx}{x\left(x + 1\right)} = - dt

so question becomes

−∫tdt=−t22+C- ∫ tdt = - \dfrac{t^{2}}{2} + C
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