`int(x-15)/((x^2+2x+5)(x^2+6x+10))dx` Please, help with this integral! I've been struggling for three days. It's partial fractions...

We are asked to find `int (x-15)/((x^2+2x+5)(x^2+6x+10))dx`

We will use partial fractions to rewrite the integrand:

`(x-15)/((x^2+2x+5)(x^2+6x+10))=(Ax+B)/(x^2+2x+5)+(Cx+D)/(x^2+6x+10)`

Now multiply both sides of the equation by the LCD to get:

`x-15=(Ax+B)(x^2+6x+10)+(Cx+D)(x^2+2x+5)`

Distributing and collecting like terms we get:

`x-15=(A+C)x^3+(6A+B+2C+D)x^2+(10A+6B+5C+2D)x+(10B+5D)`

Equating the coefficients we get A+C=0,6A+B+2C+D=0

10A+6B+5C+2D=1 and 10B+5D=-15

Solving the system yields A=B=1,C=-1,D=-5. Thus

`int (x+1)/(x^2+2x+5)dx-int (x+5)/(x^2+6x+10)dx`

The first integral can be done by letting `u=x^2+2x+5` and `du=2x+2 dx ` so

` int (x+1)/(x^2+2x+5)dx=1/2int (du)/u = 1/2 ln|x^2+2x+5| `

 Rewrite the second integral as:

`int (x+5)/(x^2+6x+10)dx = int (2x+6)/(x^2+6x+10)dx - 4int (dx)/(x^2+6x+10)`

Then we get:

`1/2 ln|x^2+6x+10|`

and 

`-4 int (dx)/((x+3)^2+1)=-4tan^(-1)(x+3)`

Thus:

`int (x-15)/((x^2+2x+5)(x^2+6x+10))dx='`

`1/2(ln|x^2+2x+5|-ln|x^2+6x+10|)-4tan^(-1)(x+3)+C