Integral of (x²/4-x/2)5(x-1)dx dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 5 \left(\frac{x^{2}}{4} - \frac{x}{2}\right)$.

      Then let $du = \left(\frac{5 x}{2} - \frac{5}{2}\right) dx$ and substitute $\frac{2 du}{5}$:

      $$\int \frac{2 u}{5}\, du$$

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

         $$\int u\, du = \frac{2 \int u\, du}{5}$$

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

            $$\int u\, du = \frac{u^{2}}{2}$$

         So, the result is: $\frac{u^{2}}{5}$

      Now substitute $u$ back in:

      $$5 \left(\frac{x^{2}}{4} - \frac{x}{2}\right)^{2}$$

   Method #2

   1. Rewrite the integrand:

      $$5 \left(\frac{x^{2}}{4} - \frac{x}{2}\right) \left(x - 1\right) = \frac{5 x^{3}}{4} - \frac{15 x^{2}}{4} + \frac{5 x}{2}$$

   2. Integrate term-by-term:

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

         $$\int \frac{5 x^{3}}{4}\, dx = \frac{5 \int x^{3}\, dx}{4}$$

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

            $$\int x^{3}\, dx = \frac{x^{4}}{4}$$

         So, the result is: $\frac{5 x^{4}}{16}$

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

         $$\int \left(- \frac{15 x^{2}}{4}\right)\, dx = - \frac{15 \int x^{2}\, dx}{4}$$

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

            $$\int x^{2}\, dx = \frac{x^{3}}{3}$$

         So, the result is: $- \frac{5 x^{3}}{4}$

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

         $$\int \frac{5 x}{2}\, dx = \frac{5 \int x\, dx}{2}$$

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

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

         So, the result is: $\frac{5 x^{2}}{4}$

      The result is: $\frac{5 x^{4}}{16} - \frac{5 x^{3}}{4} + \frac{5 x^{2}}{4}$

2. Now simplify:

   $$\frac{5 x^{2} \left(x - 2\right)^{2}}{16}$$

3. Add the constant of integration:

   $$\frac{5 x^{2} \left(x - 2\right)^{2}}{16}+ \mathrm{constant}$$

The answer is:

$$\frac{5 x^{2} \left(x - 2\right)^{2}}{16}+ \mathrm{constant}$$