∫ ( 1 + l o g e x ) 2 l o g e x d x =

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

MathematicsIndefinite IntegrationAP EAMCETAP EAMCET 2017 (26 Apr Shift 2)

Options:

A $-\frac{x}{1+\log _e x}+C$
B $\frac{x}{\left(1+\log _e x\right)^2}+C$
C $\frac{x}{\left(1+\log _e x\right)}+C$
D $\frac{-x}{\left(1+\log _e x\right)^2}+C$

Solution:

1192 Upvotes Verified Answer

The correct answer is: $\frac{x}{\left(1+\log _e x\right)}+C$

No solution. Refer to answer key.

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