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Question

5. x log 2x

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Solution

The integral is given as,

I= xlog2xdx

Use integration by parts. Consider log2xas first function and xas second function.

I= xlog2xdx =log2x xdx ( d dx log2x xdx )dx

On integrating, we get

I=log2x× x 2 2 2 2x ( x 2 2 )dx = x 2 log2x 2 1 2 xdx I= x 2 log2x 2 x 2 4 +C


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