Integral of x^3dx/(x^2-3x+2) dx

1. Rewrite the integrand:

   $$x^{3} \cdot 1 \cdot \frac{1}{x^{2} - 3 x + 2} = x + 3 - \frac{1}{x - 1} + \frac{8}{x - 2}$$

2. Integrate term-by-term:

   1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

      $$\int x\, dx = \frac{x^{2}}{2}$$

   1. The integral of a constant is the constant times the variable of integration:

      $$\int 3\, dx = 3 x$$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \left(- \frac{1}{x - 1}\right)\, dx = - \int \frac{1}{x - 1}\, dx$$

      1. Let $u = x - 1$.

         Then let $du = dx$ and substitute $du$:

         $$\int \frac{1}{u}\, du$$

         1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

         Now substitute $u$ back in:

         $$\log{\left(x - 1 \right)}$$

      So, the result is: $- \log{\left(x - 1 \right)}$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \frac{8}{x - 2}\, dx = 8 \int \frac{1}{x - 2}\, dx$$

      1. Let $u = x - 2$.

         Then let $du = dx$ and substitute $du$:

         $$\int \frac{1}{u}\, du$$

         1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

         Now substitute $u$ back in:

         $$\log{\left(x - 2 \right)}$$

      So, the result is: $8 \log{\left(x - 2 \right)}$

   The result is: $\frac{x^{2}}{2} + 3 x + 8 \log{\left(x - 2 \right)} - \log{\left(x - 1 \right)}$

3. Add the constant of integration:

   $$\frac{x^{2}}{2} + 3 x + 8 \log{\left(x - 2 \right)} - \log{\left(x - 1 \right)}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{2}}{2} + 3 x + 8 \log{\left(x - 2 \right)} - \log{\left(x - 1 \right)}+ \mathrm{constant}$$