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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=drx

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 xr ( drx ) = 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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