Question

Integrate (x^4 +2x^3 -1)/(x+1)(x+2) dx?

Answer 1

`1/3x^3-1/2x^2+x-2 log|x+1|+log|x+2|+C`

Explanation:

`x^4+2x^3-1 = x^2(x^2+3x+2)-x^3-2x^2-1 =x^2(x^2+3x+2) - x(x^2+3x+2)+x^2+2x-1 = (x^2-x)(x^2+3x+2)+1(x^2+3x+2)-x-3 = (x^2-x+1)(x^2+3x+2)-(x+3)`

Thus

`{x^4+2x*3-1}/{(x+1)(x+2)} = x^2+1-{x+3}/{(x+1)(x+2)}`

We now write

`{x+3}/{(x+1)(x+2)} = A/{x+1}+B/{x+2}`

so that

`x+3 = A(x+2) + B(x+1) = (A+B)x+(2A+B)`

This leads to

`A+B = 1, qquad 2A+B = 3 implies A=2,qquad B = -1`

so that

`int {x^4+2x*3-1}/{(x+1)(x+2)} dx`
`= int [(x^2-x+1) -2/{x+1}+1/{x+2}] dx`
`= 1/3x^3-1/2x^2+x-2 log|x+1|+log|x+2|+C`