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Question

Evaluate : 1(x+x)dx

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Solution

I=1(x+x)dx=1x(1+x)dx

Put(1+x)=t1xdx=2dt

1(x+x)dx=1x(1+x)dx=21tdt=2log|t|+C=2log(1+x)+C.

Hence, this is the answer.

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