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Find 1/k from ∫dx/(x√(1−x²⁰¹³)) | JEE

JEE Maths question with a full step-by-step solution.

Question
If ∫dxx1−x2013=k log⁡∣1−x2013−11−x2013+1∣+C\displaystyle\int \dfrac{dx}{x\sqrt{1-x^{2013}}}=k\,\log\left|\dfrac{\sqrt{1-x^{2013}}-1}{\sqrt{1-x^{2013}}+1}\right|+C, then 1k\dfrac{1}{k} equals
A20102010
B20112011
C20122012
D20132013correct
Solution
Step 1: Substitute 1−x2013=t21-x^{2013}=t^{2}. Differentiating, −2013 x2012 dx=2t dt⇒x2012 dx=−2t2013 dt-2013\,x^{2012}\,dx=2t\,dt\Rightarrow x^{2012}\,dx=-\dfrac{2t}{2013}\,dt. Step 2: Introduce x2012x^{2012} by multiplying numerator and denominator by it:
I=∫x2012 dxx20131−x2013I=\int\dfrac{x^{2012}\,dx}{x^{2013}\sqrt{1-x^{2013}}}
 Here x2013=1−t2x^{2013}=1-t^{2} and 1−x2013=t\sqrt{1-x^{2013}}=t. Step 3: Substitute:
I=∫−2t2013 dt(1−t2) t=−22013∫dt1−t2I=\int\dfrac{-\tfrac{2t}{2013}\,dt}{(1-t^{2})\,t}=-\dfrac{2}{2013}\int\dfrac{dt}{1-t^{2}}
 Step 4: Use ∫dt1−t2=12ln⁡∣1+t1−t∣\displaystyle\int\dfrac{dt}{1-t^{2}}=\dfrac{1}{2}\ln\left|\dfrac{1+t}{1-t}\right|:
I=−22013⋅12ln⁡∣1+t1−t∣+C=12013ln⁡∣1−t1+t∣+CI=-\dfrac{2}{2013}\cdot\dfrac{1}{2}\ln\left|\dfrac{1+t}{1-t}\right|+C=\dfrac{1}{2013}\ln\left|\dfrac{1-t}{1+t}\right|+C
 Step 5: Back-substitute t=1−x2013t=\sqrt{1-x^{2013}}; since ∣1−t1+t∣=∣t−1t+1∣\left|\dfrac{1-t}{1+t}\right|=\left|\dfrac{t-1}{t+1}\right|,
I=12013log⁡∣1−x2013−11−x2013+1∣+C  ⇒  k=12013, 1k=2013I=\dfrac{1}{2013}\log\left|\dfrac{\sqrt{1-x^{2013}}-1}{\sqrt{1-x^{2013}}+1}\right|+C\;\Rightarrow\; k=\dfrac{1}{2013},\ \dfrac{1}{k}=2013
 Correct answer: (4)
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