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Question

Integrate with respect to x :
1(x+1)(x+2)

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Solution

1(x+1)(x+2)dx
BY partial fraction method;
1(x+1)(x+2)=A(x+1)+B(x+2)1=A(x+2)+B(x+1)1=(A+B)x+(2A+B)
Comparing both side we get,
A+B=0and2A+B=1A=1andB=11(x+1)(x+2)dx=[1(x+1)1(x+2)]dx=log(x+1)log(x+2)+C=log(x+1x+2)+C

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