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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 ∫ (x - 4) x^ 2 x - 2 dx is equal to (where, C is the...

The value of ∫(x−4)x2x−2dx∫ \dfrac{\left(x - 4\right)}{x^{2}\sqrt{x - 2}}dx is equal to (where, CC is the constant of integration)

Options

  1. A2xx−2+C2x\sqrt{x - 2} + C
  2. B−2xx−2+C- \dfrac{2}{x}\sqrt{x - 2} + C
  3. Cx−2x+C\dfrac{\sqrt{x - 2}}{x} + C
  4. Dxx−2+C\dfrac{x}{\sqrt{x - 2}} + C

Correct answer

B. −2xx−2+C- \dfrac{2}{x}\sqrt{x - 2} + C

Step-by-step solution

∫x−4x2x−2dx∫ \dfrac{x - 4}{x^{2}\sqrt{x - 2}}dx
∫(1x2−4x3)1x−2x2dx∫ \dfrac{\left(\dfrac{1}{x^{2}} - \dfrac{4}{x^{3}}\right)}{\sqrt{\dfrac{1}{x} - \dfrac{2}{x^{2}}}}dx
Let 1x−2x2=t\dfrac{1}{x} - \dfrac{2}{x^{2}} = t
\left(- \dfrac{1}{x2} + \dfrac{4}{x3}\right)dx = dt<br/><br />∫ \dfrac{- dt}{t} = - 2\sqrt{t} + C<br/><br />= - 2\sqrt{\dfrac{1}{x} - \dfrac{2}{x^{2}}} + C<br/><br />= - \dfrac{2}{x}\sqrt{x - 2} + C
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