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

Explanation

Let I = `int(x^2+x+1)/((x+2)(x^2+1)`dx
considering partial fraction we can write it as
`int(x^2+x+1)/((x+2)(x^2+1)`dx = `A/(x+2) + (Bx+c)/(x^2 +1)`
Multiply both side by `(x+2)(x^2+1)`
we have `x^2 +x+1 = A( x^2+1) +(Bx+c)(x+2)`
put `x+2=0` or `x=-2` we have `A= 3/5`

comparing the coefficients of `x^2` on both sides we have `1 = A+B`
=> `1= 3/5 +B`
=> `B=2/5`
comparing the coefficients of `x` on both sides we have `1 = 2B+C`
`1= 2(2/5) +C`
`C= 1/5`

so `(x^2+x+1)/((x+2)(x^2+1)`dx =`(3/5)/(x+2)` +`((2/5)x +1/5)/(x^2+1)`
now
`I= 3/5 int 1/(x+2) dx + 1/5 int (2x+1)/(x^2+1)`
`I= 3/5 int 1/(x+2) dx + 1/5 int ((2x)/(x^2+1) + 1/ (x^2+1))` dx
= `3/5 log|x+2| + 1/5 { log|x^2+1| + tan^-1(x)|} +c`