Question

How do you integrate `int x^2/ (x^2+x+4)` using partial fractions?

Answer 1

`I=x-1/2ln|x^2+x+4|-7/sqrt15 tan^-1((2x+1)/sqrt15)+c`

Explanation:

We know that,

`color(red)((1)int(d(f(x)))/f(x) =ln|f(x)|+c`

`color(blue)((2)int1/(x^2+a^2)dx=1/atan^-1(x/a)+c`

Here,

`I=intx^2/(x^2+x+4)dx`

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

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

`=int1dx-1/2int(2x+8)/(x^2+x+4)dx`

`=x-1/2int((2x+1)+7)/(x^2+x+4)dx`

`=x-1/2int(2x+1)/(x^2+x+4)dx-1/2int7/(x^2+x+4)dx`

#=x-1/2int(color(red)(d(x^2+x+4))/color(red)((x^2+x+4)))dx- 7/2int1/(x^2+x+1/4+15/4)dx#

#=x-1/2color(red)(ln|x^2+x+4|)-7/2color(blue)(int 1/((x+1/2)^2+ (sqrt15/2)^2)dx#

#=x-1/2ln|x^2+x+4|-7/2*color(blue)(1/(sqrt15/2) tan^-1((x+1/2)/(sqrt15/2)))+c#

`=x-1/2ln|x^2+x+4|-7/sqrt15 tan^-1((2x+1)/sqrt15)+c`