This is the indefinite integral I have to evaluate: ∫x3x2−x−2dx
so by using the long division on polynomials technique, I got to: x22+x+∫3x+2x2−x−2dx
How do I continue from here? I thought of using "integration of rational functions" but it didn't work.
Answer
Specifically, you have that 3x+2x2−x−2=Ax−2+Bx+1
The usual technique reveals that A=83 and B=13
Hence x3x2−x−2=x+1+3x+2(x−2)(x+1)=x+1+83(x−2)+13(x+1)
So that ∫x3x2−x−2dx=∫x+1+3x+2x2−x−2=∫x+1+83(x−2)+13(x+1)dx
Hence ∫x3x2−x−2=x+x22+83ln|x−2|+13ln|x+1|+C
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