Question

Partial fractions `\int5/((x^2+2x+2)(x-1))`?

I am getting results like this, are they ok so far?

`A+C=0`, `B-A+2C=0`, `2C-B=5`

Answer 1

The answer is `=ln(|x-1|)-1/2ln(|x^2+2x+2|)-2arctan(x+1)+C`

Explanation:

We need

`int(u'(x)dx)/(u(x))=ln(u(x))+C`

`int(dx)/(x^2+1)=arctan(x)+C`

Perform the decomposition into partial fractions

`(5)/((x-1)(x^2+2x+2))=A/(x-1)+(Bx+C)/(x^2+2x+2)`

`=(A(x^2+2x+2)+(Bx+C)(x-1))/((x-1)(x^2+2x+2))`

The denominatoras are the same, compare the numerators

`5=A(x^2+2x+2)+(Bx+C)(x-1)`

Let `x=1`, `=>`, `5=5A`, `=>`, `A=1`

Let `x=0`, `=>`, `5=2A-C`, `=>`, `C=2A-5=2-5=-3`

Coefficients of `x^2`,

`0=A+B`, `=>`, `B=-A=-1`

Therefore,

`(5)/((x-1)(x^2+2x+2))=1/(x-1)+(-x-3)/(x^2+2x+2)`

So,

`int(5dx)/((x-1)(x^2+2x+2))=int(1dx)/(x-1)-int((x+3)dx)/(x^2+2x+2)`

`=ln(|x-1|)-I`

And

`I=int((x+3)dx)/(x^2+2x+2)=int((1/2(2x+2)+2)dx)/(x^2+2x+2)`

`=1/2int((2x+2)dx)/(x^2+2x+2)+2int(dx)/(x^2+2x+2)`

`=1/2ln(|x^2+2x+2|)+2I_1`

`I_1=int(dx)/(x^2+2x+2)=int(dx)/((x+1)^2+1)`

`=arctan(x+1)`

Finally,

`int(5dx)/((x-1)(x^2+2x+2))=ln(|x-1|)-1/2ln(|x^2+2x+2|)-2arctan(x+1)+C`