∫dx(x2+x+1)2
=∫dx((x+12)2+34)2
Let x+12=√32tanθ
⇒dx=√32sec2θ
∴∫√32.sec2θ916sec4θdθ
=83√3∫cos2θdθ
=43√3∫2cos2θdθ
=43√3∫(1+cos2θ)dθ
=43√3(θ+sin2θ2)+C
Now, tanθ=2x+1√3
⇒sin2θ=2(sinθ)(cosθ)
⇒sin2θ=√3(2x+1)2(x2+x+1)
∴43√3(tan−1(2x+1√3)+√34(2x+1x2+x+1))+C
=43√3tan−1(2x+1√3)+13(2x+1x2+x+1)+C
⇒a=43√3,b=13
∴9(√3a+b)=15