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JEE Main 20146 April
Exam
JEE Main
Year
2014
Date
6 April
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2014 · 6 April · Mathematics · Indefinite Integration

The integral ∫ (1 + x - (1)/(x)) e ^ x + (1)/(x) dx , is equal to

The integral ∫(1+x−1x)ex+1xdx∫ \left(1 + x - \dfrac{1}{x}\right){e}^{x + \dfrac{1}{x}}dx, is equal to

Options

  1. A(x+1)ex+1x+c\left(x + 1\right){e}^{x + \dfrac{1}{x}} + c
  2. B−xex+1x+c- x{e}^{x + \dfrac{1}{x}} + c
  3. C(x−1)ex+1x+c\left(x - 1\right){e}^{x + \dfrac{1}{x}} + c
  4. Dxex+1x+cx{e}^{x + \dfrac{1}{x}} + c

Correct answer

D. xex+1x+cx{e}^{x + \dfrac{1}{x}} + c

Step-by-step solution

∫(1+x−1x)ex+1xdx=∫ex+1xdx+∫x(1−1x2)ex+1xdx∫ \left(1 + x - \dfrac{1}{x}\right){e}^{x + \dfrac{1}{x}}dx = ∫ {e}^{x + \dfrac{1}{x}}dx + ∫ x\left(1 - \dfrac{1}{{x}^{2}}\right){e}^{x + \dfrac{1}{x}}dx

Integrating ∫x⏟I.(1−1x2)ex+1x⏟IIdx∫ \mathop{\mathop{x}_{⏟}}_{I} . \mathop{\mathop{( 1 - \dfrac{1}{{x}^{2}} ){e}^{x + \dfrac{1}{x}}}_{⏟}}_{II}dx, by parts

I=xex+1x−∫ex+1xdx+cI = x{e}^{x + \dfrac{1}{x}} - ∫ {e}^{x + \dfrac{1}{x}}dx + c

=∫(1+x−1x)ex+1xdx=∫ex+1xdx+xex+1x−∫ex+1xdx+c= ∫ \left(1 + x - \dfrac{1}{x}\right){e}^{x + \dfrac{1}{x}}dx = ∫ {e}^{x + \dfrac{1}{x}}dx + x{e}^{x + \dfrac{1}{x}} - ∫ {e}^{x + \dfrac{1}{x}}dx + c

=xex+1x+c= x{e}^{x + \dfrac{1}{x}} + c, where cc is the constant of integration.

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