# what is the int. of (x^2 + x + 1)^5 dx?

### 1 Answer | Add Yours

You should consider `(x^2+x+1)^5 = (x^2+x+1)^2*(x^2+x+1)^2*(x^2+x+1)`

Expanding the square of trinomial yields:

`(x^2+x+1)^2 = x^4 + x^2 + 1 + 2x^3 + 2x^2 + 2x`

`(x^2+x+1)^2 = x^4 + 2x^3 + 3x^2 + 2x + 1`

You need to multiply `(x^2+x+1)^2` by `(x^2+x+1)^2` such that:

`(x^4 + 2x^3 + 3x^2 + 2x + 1)(x^4 + 2x^3 + 3x^2 + 2x + 1) = (x^2+x+1)^4`

Collecting like terms yields:

`(x^2+x+1)^4 = x^8 + 4x^7 + 10x^6 + 14x^5 + 21x^4 + 16x^3 + 10x^2 + 4x + 1`

You need to multiply `(x^2+x+1)^4` by `(x^2+x+1)` such that:

`(x^2+x+1)^5 = (x^8 + 4x^7 + 10x^6 + 14x^5 + 21x^4 + 16x^3 + 10x^2 + 4x + 1)(x^2+x+1)`

`(x^2+x+1)^5 = x^10 + x^9 + x^8 + 4x^9 + 4x^8 + 4x^7 + 10x^8 + 10x^7 + 10x^6 + 14x^7 + 14x^6 + 14x^5 + 21x^6 + 21x^5 + 21x^4 + 16x^5 + 16x^4 + 16x^3 + 10x^4 + 10x^3 + 10x^2 + 4x^3 + 4x^2 + 4x + x^2 + x + 1`

`(x^2+x+1)^5 = x^10 + 5x^9 + 15x^8 + 28x^7 + 45x^6 + 51x^5 + 47x^4 + 30x^3 + 15x^2 + 5x + 1`

You need to evaluate the integral of `(x^2+x+1)^5,` hence, you may split the integra into more simpler integrals such that:

`int (x^2+x+1)^5 dx = int x^10 dx + int 5x^9 dx + int 15x^8 dx + int 28x^7 dx + int 45x^6 dx + int 51x^5 dx + int 47x^4 dx + int 30x^3 dx + int 15x^2 dx + int 5x dx + int 1 dx`

`int (x^2+x+1)^5 dx = (x^11)/11 + 5(x^10)/10 + 15(x^9)/9 + 28(x^8)/8 + 45(x^7)/7 + 51(x^6)/6 + 47(x^5)/5 + 30(x^4)/4 + 15(x^3)/3 + 5(x^2)/2 + x + c`

**Hence, evaluating the given integral yields** `int (x^2+x+1)^5 dx = (x^11)/11 + 5(x^10)/10 + 15(x^9)/9 + 28(x^8)/8 + 45(x^7)/7 + 51(x^6)/6 + 47(x^5)/5 + 30(x^4)/4 + 15(x^3)/3 + 5(x^2)/2 + x + c`

**Sources:**