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Question

Evaluate the definite integrals.
206x+3x2+4dx.


Solution

Let I=206x+3x2+4dx=206xx2+4dx+203x2+4dx
Put x2+4=t2x=dtdxdx=dt2x
Lower limit, when x =0, t =0 +4 =4
upper limit, when x =2, t =4+4 =8
I=846xtdt2x+203x2+4dx=3841tdt+3201x2+22dx=3[logt]84+32[tan1x2]20[dxa+x2=1atan1xa]=3[log(8)log(4)+32][tan122]=3log(84)+32×π4     [logbloga=logba]=3log2+3π8

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