NIMCET 2016 Question Paper with solutions

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Question : 59 of 120

Marks: +1, -0

∫

\{(\logx-1)/{1+(\logx)^2}\}^2

dx is equal to ;

${xe^x}/{1+x^2}$ + C
$x/{(logx)^2+1}$ + C
${logx}/{(logx)^2+1}$ + C
$x/{x^2+1}$ + C

Solution:

Given

l=∫\{(\log x-1)/{1+(\log x)^2}\}^2\dx

Let log x = t

⇒x=e^t

⇒\dx=e^t\;\dt

⇒l=∫e^t(t-1)^2/(t^2+1)^2\dt

⇒I=∫e^t\;{t^2+1-2t}/(t^2+1)^2\dt

⇒l=∫e^t[1/{t^2+1}+(-2t)/(t^2+1)^2]\dt

l=e^t/{t^2+1}+c

[∵∫e^x[f(x)+f'(x)]\dx=e^x+f(x)+C]

⇒l=x/{(\log x)^2+1}+C

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