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Question

Evaluate :
1(x+3)x+2dx, on xI(2,)

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Solution

Let I=1(x+3)x+2dx

Substitute x+2=t2dx=2tdt

So, I=2tdt(t22+3)t=2dt1+t2

I=2tan1t+C=2tan1(x+2)+C

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