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Question

Integrate the function.
x2+4x+6dx.

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Solution

Let I=x2+4x+6dx=x2+4x+22+64dx
=(x+2)2+(2)2dx[x2+a2dx=x2x2+a2+a22log|x+x2+a2|]=x+22x2+4x+6+22log|(x+2)+x2+4x+6|+CI=x+22x2+4x+6+log|(x+2)+x2+4x+6|+C


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