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JEE Advanced 2018Paper 2
Exam
JEE Advanced
Year
2018
Paper
Paper 2
Subject
Mathematics
Chapter
Definite Integration

JEE Advanced 2018 · Paper 2 · Mathematics · Definite Integration

The value of the integral ∫ _ 0 ^ (1)/(2) 1 + 3 ( (x + 1) ^ 2 (1 -...

The value of the integral ∫0121+3((x+1)2(1−x)6)14dx\mathop{∫}_{0}^{\dfrac{1}{2}}\dfrac{1 + \sqrt{3}}{\left({\left(x + 1\right)}^{2}{\left(1 - x\right)}^{6}\right)\dfrac{1}{4}}dx is

Correct answer

0

Step-by-step solution

∫012(1+3)dx[(1+x)2(1−x)6]14\mathop{∫}_{0}^{\dfrac{1}{2}}\dfrac{\left(1 + \sqrt{3}\right)dx}{\left[{\left(1 + x\right)}^{2}{\left(1 - x\right)}^{6}\right]\dfrac{1}{4}}
∫012(1+3)dx(1+x)2[(1−x)6(1+x)6]14\mathop{∫}_{0}^{\dfrac{1}{2}}\dfrac{\left(1 + \sqrt{3}\right)dx}{{\left(1 + x\right)}^{2}\left[\dfrac{{\left(1 - x\right)}^{6}}{{\left(1 + x\right)}^{6}}\right]\dfrac{1}{4}}
Put 1−x1+x=t⇒−2dx(1+x)2=dt\dfrac{1 - x}{1 + x} = t ⇒ \dfrac{- 2dx}{{\left(1 + x\right)}^{2}} = dt
I=∫113(1+3)dt−2t64=−(1+3)2×∣−2t∣113=(1+3)(3−1)=2I = \mathop{∫}_{1}^{\dfrac{1}{3}}\dfrac{\left(1 + \sqrt{3}\right)dt}{- 2t\dfrac{6}{4}} = \dfrac{- \left(1 + \sqrt{3}\right)}{2} \times {\left|- \dfrac{2}{\sqrt{t}}\right|}_{1}^{\dfrac{1}{3}} = \left(1 + \sqrt{3}\right)\left(\sqrt{3} - 1\right) = 2
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