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Question

If dx(x2+x+1)2=atan1(2x+13)+b(2x+1x2+x+1)+C, x>0 where C is the constant of integration, then the value of 9(3a+b) is equal to

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Solution

dx(x2+x+1)2
=dx((x+12)2+34)2
Let x+12=32tanθ
dx=32sec2θ

32.sec2θ916sec4θdθ
=833cos2θdθ
=4332cos2θdθ
=433(1+cos2θ)dθ
=433(θ+sin2θ2)+C
Now, tanθ=2x+13
sin2θ=2(sinθ)(cosθ)
sin2θ=3(2x+1)2(x2+x+1)

433(tan1(2x+13)+34(2x+1x2+x+1))+C
=433tan1(2x+13)+13(2x+1x2+x+1)+C
a=433,b=13
9(3a+b)=15

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