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Question

Evaluate dxx32(1+1x)

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Solution

Let I=dxx32(1+x12)

Let t=1+x12dt=12x121dx

dt=12x32dx=12dxx32

I=2dtt

=2log|t|+c where c is the constant of integration.

=2log1+x12+c

=2log1+1x+c

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