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Question

Solve 13x1x2+x+1dx

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Solution

=13x1x2+x+1dx
13⎢ ⎢ ⎢x+12(x+12)2+(3/2)2dx32dx(x+1/2)2+(3/22)⎥ ⎥ ⎥
Put x+1/2=t dx=dt
=13[tdtt2+(3/2)232dtt2+(3/2)2]
Let put t2=u 2tdt=du
=13[du4+3/432×23tan1t×23]
=13[12log|(x+1/2)2+3/4|13tan1(2x+13)]+c
=16log|x2+x+1|133tan1(2x+13)+c


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