Question

# 3.x+x logx

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Solution

## The function is, y= ∫ 1 x+xlogx dx (1) Let 1+logx=r Differentiate with respect to x, dr dx = 1 x dx= dr⋅x Substitute the value of dx in equation (1) and use the formula of ∫ 1 x dx=log| x | +A, where E is constant. ∫ 1 x+xlogx dx = ∫ 1 x⋅r ⋅( dr⋅x ) = ∫ 1 r  dr =log| r |+D (2) Where, D is constant. Substitute the value of r in equation (2). y=( log| 1+logx | )+D Thus, the value of integration is ( log| 1+logx | )+D.

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