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Question

Evaluate the following integrals:

x2+x+1x2+1x+2dx

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Solution

Let I=x2+x+1x2+1x+2dxWe expressx2+x+1x2+1x+2=Ax+2+Bx+Cx2+1x2+x+1=Ax2+1+Bx+Cx+2Equating the coefficients of x2, x and constants, we get1=A+B and 1=2B+C and 1=A+2Cor A=35 and B=25 and C=15 I=35x+2+25x+15x2+1dx =351x+2dx+25xx2+1 dx+151x2+1 dx =35I1+25I2+15I3 ...(1)Now, I1=1x+2dx Let x+2=u On differentiating both sides, we get dx=du I1=1udu =logu+c1 =logx+2+c1 ...(2)And, I2=xx2+1 dx Let x2+1=u On differentiating both sides, we get 2x dx=du I2=121udu =12logu+c2 =12logx2+1+c2 ...(3)And, I3=1x2+1 dx =tan-1x+c3 ...(4)From (1), (2), (3) and (4), we get I=35logx+2+c1+2512logx2+1+c2+15tan-1x+c3 =35logx+2+15logx2+1+15tan-1x+cHence, x2+x+1x2+1x+2dx=35logx+2+15logx2+1+15tan-1x+c

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