∫ [ 1 + ( l o g x ) 2 l o g x − 1 ] 2 d x =

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Question: Answered & Verified by Expert

$\int\left[\frac{\log x-1}{1+(\log x)^{2}}\right]^{2} d x=$

MathematicsIndefinite IntegrationMHT CETMHT CET 2020 (16 Oct Shift 2)

Options:

A $\frac{x}{(1+\log x)}+c$
B $\frac{x}{1+(\log x)^{2}}+c$
C $\frac{x^{2}}{1+(\log x)^{2}}+c$
D $\frac{1}{1+(\log x)^{2}}+c$

Solution:

1033 Upvotes Verified Answer

The correct answer is: $\frac{x}{1+(\log x)^{2}}+c$

(B)

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