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Question

Evaluate: 2x2+1x2(x2+4)dx.

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Solution

Let I=2x2+1x2(x2+4)dx
Perform partial fraction decomposition:
2x2+1x2(x2+4)=Ax+Bx2+Cx+Dx2+4

2x2+1x2(x2+4)=x2(Cx+D)+x(x2+4)A+(x2+4)Bx2(x2+4)
After simplifying, we get
Coefficients near like terms should be equal, so the following system is obtained:
A+C=0,B+D=2,4A=0,4B=1.
Solve it, then A = 0, B=14, C = 0, D=74
2x2+1=x3(A+C)+x2(B+D)+4xA+4B
Therefore,
2x2+1x2(x2+4)=141x2+74(x2+4)
Integrating separately, we get
=141x2dx+741x2+4
We know that, 1x2+1=tan1x
=14(x2+12+1)+74×2(tan1(x2))+C
=14(1x)+78(tan1(x2))+C
=1x[14+78xtan1(x2)]+C

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