Question

What is the integral of `(x^3+x^2-5x+15)/((x^2+5)(x^2+2x+3))dx`?

Answer 1

`I=x^4/4+x^3/3-(5x^2)/2+15x+15ln|x^2+5|-(30tan^-1 (x/sqrt5))/sqrt5+c`

Explanation:

`I=int(x^3+x^2-5x+15/(x^2+5)(x^2+2x+3))dx`
`=int(x^3+x^2-5x)dx+15int(x^2+2x+3)/(x^2+5)dx`
`=[x^4/4+x^3/3-(5x^2)/2]+15int(x^2+5+2x-2)/(x^2+5)dx`
`=[x^4/4+x^3/3-(5x^2)/2]+15int(1+(2x)/(x^2+5)-(2)/(x^2+5))dx`
`=x^4/4+x^3/3-(5x^2)/2+15x+15ln|x^2+5|-30int1/(x^2+(sqrt5)^2)dx`
`=x^4/4+x^3/3-(5x^2)/2+15x+15ln(x^2+5)-30(1/sqrt5)tan^-1(x/sqrt5)+c`
`=x^4/4+x^3/3-(5x^2)/2+15x+15ln(x^2+5)-(30tan^-1 (x/sqrt5))/sqrt5+c`

Answer 2

`int \ (x^3+x^2-5x+15)/((x^2+5)(x^2+2x+3)) \ dx = (ln|x^2+2x+3|)/2 + 5 sqrt(2)/2 arctan((x+1)/sqrt(2)) - sqrt(5) arctan(x/sqrt(5)) + C`

Explanation:

We seek:

`I = int \ (x^3+x^2-5x+15)/((x^2+5)(x^2+2x+3)) \ dx`

We can decompose the integrand into partial fraction, which will be of the form:

`(x^3+x^2-5x+15)/((x^2+5)(x^2+2x+3)) -= (Ax+B)/(x^2+5) + (Cx+D)/(x^2+2x+3)`

`" " = ((Ax+B)(x^2+2x+3) + (Cx+D)(x^2+5))/((x^2+5)(x^2+2x+3))`

Leading to an Identity:

`x^3+x^2-5x+15 -= (Ax+B)(x^2+2x+3) + (Cx+D)(x^2+5)`

Where `A,B,C,D` are constants to be determined. We can compare coefficients:

`Coef(x^3) : 1 = A + C`
`Coef(x^2) : 1 = 2A+B+D`
`Coef(x^1) : -5 = 3A+2B+5C`
`Coef(x^0) : 15 = 3B+5D`

If we solve these we get:

`A=0, B=-5, C=1, D=6`

Thus we have:

`I = int \ (x+6)/(x^2+2x+3) - (5)/(x^2+5) \ dx`

`\ \ = 1/2 int \ (2x+12)/(x^2+2x+3) \ dx - int \ (5)/(x^2+5) \ dx`

`\ \ = 1/2 int \ ((2x+2) + 10)/(x^2+2x+3) \ dx - int \ (5)/(x^2+5) \ dx`

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

`\ \ = 1/2 I_1 + 5 I_2- 5I_3 + c`, say

Where we now consider each of the three integrals individually:

`I_1 = int \ (2x+2)/(x^2+2x+3) \ dx`

We can perform a substitution, `u=x^2+2x+3 => (du)/dx=2x+2` so:

`I_1 = int \ 1/u \ du = ln |u| = ln|x^2+2x+3|`

`I_2= int \ 1/(x^2+2x+3) = int \ 1/((x+1)^2-1^2+3) = int \ 1/((x+1)^2+2)`

We can perform a substitution, `usqrt(2)=x+1 => sqrt(2)(du)/dx=1` so:

`I_2= int \ 1/((usqrt(2))^2+2) \ sqrt(2) \ du`
`\ \ \ = int \ 1/(2u^2+2) \ sqrt(2) \ du`
`\ \ \ = sqrt(2)/2 \ int \ 1/(u^2+1) \ du`
`\ \ \ = sqrt(2)/2 \ arctan(u)`
`\ \ \ = sqrt(2)/2 \ arctan((x+1)/sqrt(2))`

Finally:

`I_3 int \ (1)/(x^2+5) \ dx`
We can perform a substitution, `usqrt(5)=x => sqrt(5)(du)/dx=1` so:
`I_3 = int \ (1)/((usqrt(5))^2+5) \ sqrt(5) \ du`
`\ \ \ = int \ (1)/(5u^2+5) \ sqrt(5) \ du`
`\ \ \ = sqrt(5)/5 int \ (1)/(u^2+1) \ du`
`\ \ \ = sqrt(5)/5 arctan(u)`
`\ \ \ = sqrt(5)/5 arctan(x/sqrt(5))`

Combining all three results:

`I = 1/2 (ln|x^2+2x+3|) + 5 (sqrt(2)/2 \ arctan((x+1)/sqrt(2))) - 5(sqrt(5)/5) arctan(x/sqrt(5))`

`\ \ \ = (ln|x^2+2x+3|)/2 + 5 sqrt(2)/2 arctan((x+1)/sqrt(2)) - sqrt(5) arctan(x/sqrt(5)) + C`