Question

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

Answer 1

`I=x-1/2ln|x^2+x+2|-3/sqrt7tan^-1((2x+1)/sqrt7)+c`

Explanation:

We know that,

`color(red)((1)int(f'(x))/(f(x))dx=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+2)dx`

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

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

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

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

`=x-1/2int(2x+1)/(x^2+x+2)dx-1/2int3/(x^2+x+2)dx`

#=x-1/2color(red)(int(d/(dx) (x^2+x+2))/(x^2+x+2)dx)-3/2int1/(x^2+x+1/4+7/4)dx#

#=x-1/2color(red)(ln|x^2+x+2|)-3/2color(blue)(int1/((x+1/2)^2+ (sqrt7/2)^2)dx#

#=x-1/2ln|x^2+x+2|-3/2*color(blue) (1/(sqrt7/2)tan^-1((x+1/2)/(sqrt7/2))+C#

`=x-1/2ln|x^2+x+2|-3/sqrt7tan^-1((2x+1)/sqrt7)+C`