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JEE Main 201610 AprilOnline
Exam
JEE Main
Year
2016
Date
10 April
Mode
Online
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2016 · 10 April · Online · Mathematics · Indefinite Integration

The integral ∫ dx (1 + x ) x - x ^ 2 is equal to

The integral ∫ \dfrac{dx}{\left(1 + \sqrt{x}\right)\sqrt{x - {x}^{2}}} is equal to

Options

  1. A−21+x1−x+c- 2\sqrt{\dfrac{1 + \sqrt{x}}{1 - \sqrt{x}}} + c
  2. B−1−x1+x+c- \sqrt{\dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}} + c
  3. C−21−x1+x+c- 2\sqrt{\dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}} + c
  4. D1+x1−x+c\sqrt{\dfrac{1 + \sqrt{x}}{1 - \sqrt{x}}} + c

Correct answer

C. −21−x1+x+c- 2\sqrt{\dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}} + c

Step-by-step solution

I=∫dx(1+x)x1−xI = ∫ \dfrac{dx}{\left(1 + \sqrt{x}\right)\sqrt{x}\sqrt{1 - x}}

Put, 1+x=t⇒12xdx=dt1 + \sqrt{x} = t ⇒ \dfrac{1}{2\sqrt{x}}dx = dt

And, 1−x=1−(t−1)2=2t−t21 - x = 1 - {\left(t - 1\right)}^{2} = 2t - {t}^{2}

∴I=∫2dtt2t−t2∴ I = ∫ \dfrac{2dt}{t\sqrt{2t - {t}^{2}}}

Again put, t=1z⇒dt=−1z2dzt = \dfrac{1}{z} ⇒ dt = \dfrac{- 1}{{z}^{2}}dz

⇒I=2∫−1z2dz1z2z−1z2=2∫−dz2z−1=−22z−1+c⇒ I = 2 ∫ \dfrac{- \dfrac{1}{{z}^{2}}dz}{\dfrac{1}{z}\sqrt{\dfrac{2}{z} - \dfrac{1}{{z}^{2}}}} = 2 ∫ \dfrac{- dz}{\sqrt{2z - 1}} = - 2\sqrt{2z - 1} + c

Where cc is arbitrary constant.

=−22t−1+c= - 2\sqrt{\dfrac{2}{t} - 1} + c

=−22−tt+c= - 2\sqrt{\dfrac{2 - t}{t}} + c

=−21−x1+x+c= - 2\sqrt{\dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}} + c

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