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Question

Evaluate the integrals:
x4(x2+1)(x1)dx

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Solution

=(x41+1)dx(x2+1)(x1)
=(x41)dx(x2+1)(x1)+dx(x2+1)(x1)
=(x21)(x2+1)dx(x2+1)(x1)+dx(x2+1)(x1)
=(x21)dx(x1)+dx(x2+1)(x1)
=(x1)(x+1)dx(x1)+dx(x2+1)(x1)
=(x+1)dx+dx(x2+1)(x1)
Consider dx(x2+1)(x1)=Ax1+Bx+Cx2+1
1=A(x2+1)+B(x1)
Put x=11=2A or A=12
Put x=01=AC or C=A1=121=12
Put x=11=2A+(B+C)(2)
or 2A+2B2C=1
or 2×12+2B2×12=1
or 2B=1 or B=12
(x+1)dx+12dxx1+12x12x2+1dx
=(x+1)dx+12dxx1+12x+1x2+1dx
=(x+1)dx+12dxx1+142x+2x2+1dx
=(x+1)dx+12dxx1+142xx2+1dx+142x2+1dx
=(x+1)dx+12dxx1+142xx2+1dx+12dxx2+1
=x22+x+12log(x1)14log(x2+1)12tan1x+c where c is the constant of integration.

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