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Find 1/k from ∫dx/(x√(1−x²⁰¹³)) | JEE

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

Question

\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

\dfrac{1}{k}

equals

2010

2011

2012

2013

correct

Solution

Step 1: Substitute

1-x^{2013}=t^{2}

. Differentiating,

-2013\,x^{2012}\,dx=2t\,dt\Rightarrow x^{2012}\,dx=-\dfrac{2t}{2013}\,dt

. Step 2: Introduce

x^{2012}

by multiplying numerator and denominator by it:

I=\int\dfrac{x^{2012}\,dx}{x^{2013}\sqrt{1-x^{2013}}}

Here

x^{2013}=1-t^{2}

and

\sqrt{1-x^{2013}}=t

. Step 3: Substitute:

I=\int\dfrac{-\tfrac{2t}{2013}\,dt}{(1-t^{2})\,t}=-\dfrac{2}{2013}\int\dfrac{dt}{1-t^{2}}

Step 4: Use

\displaystyle\int\dfrac{dt}{1-t^{2}}=\dfrac{1}{2}\ln\left|\dfrac{1+t}{1-t}\right|

I=-\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=\sqrt{1-x^{2013}}

; since

\left|\dfrac{1-t}{1+t}\right|=\left|\dfrac{t-1}{t+1}\right|

I=\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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