The integral of f(x) = (x^2 - 2 x +3)^2 (x+1) dx is

[tex]\displaystyle \int \left(x^2-2x+3\right)^2(x+1) \, \mathrm dx =\\= \int \left(x^4-2x^3+3x^2-2x^3+4x^2-6x+3x^2-6x+9\right)(x+1) \, \mathrm dx =\\=\int \left(x^4-4x^3+10x^2-12x+9\right)(x+1) \, \mathrm dx=\\=\int x^5-4x^4+10x^3-12x^2+9x+x^4-4x^3+10x^2-12x+9 \, \mathrm dx=\\=\int x^5-3x^4+6x^3-2x^2-3x+9 \, \mathrm dx[/tex]

Use the sum rule: [tex]\displaystyle \int{f(x)+g(x)}\,\mathrm dx=\int f(x)\,\mathrm dx + \int g(x)\,\mathrm dx[/tex].

[tex]\displaystyle \int x^5-3x^4+6x^3-2x^2-3x+9 \, \mathrm dx= \\= \int x^5\,\mathrm dx -\int 3x^4 \,\mathrm dx+\int 6x^3\,\mathrm dx-\int 2x^2\,\mathrm dx-\int 3x\,\mathrm dx+\int9 \,\mathrm dx[/tex]

Use the constant rule: [tex]\displaystyle \int cf(x)\, \mathrm dx=c\int f(x)\,\mathrm dx[/tex]

[tex]\displaystyle\int x^5\,\mathrm dx -\int 3x^4 \,\mathrm dx+\int 6x^3\,\mathrm dx-\int 2x^2\,\mathrm dx-\int 3x\,\mathrm dx+\int9 \,\mathrm dx=\\=\int x^5\,\mathrm dx -3\int x^4 \,\mathrm dx+6\int x^3\,\mathrm dx-2\int x^2\,\mathrm dx-3\int x\,\mathrm dx+\int9 \,\mathrm dx[/tex]

The integral of a constant: [tex]\displaystyle \int{a}\, \mathrm dx=ax+C[/tex] and

The power rule: [tex]\displaystyle \int{x^n} \mathrm dx=\dfrac{x^{n+1}}{n+1}[/tex]

[tex]\displaystyle\int x^5\,\mathrm dx -3\int x^4 \,\mathrm dx+6\int x^3\,\mathrm dx-2\int x^2\,\mathrm dx-3\int x\,\mathrm dx+\int9 \,\mathrm dx=\\\\=\dfrac{x^6}{6}-\dfrac{3x^5}{5}+\dfrac{6x^4}{4}-\dfrac{2x^3}{3}-\dfrac{3x^2}{2}+9x+C=\boxed{\dfrac{x^6}{6}-\dfrac{3x^5}{5}+\dfrac{3x^4}{2}-\dfrac{2x^3}{3}-\dfrac{3x^2}{2}+9x+C}[/tex]

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