Integral of 10x^3*9x^2 dx

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

   $$\int 10 x^{3} \cdot 9 x^{2}\, dx = 90 \int x^{2} x^{3}\, dx$$

   1. There are multiple ways to do this integral.

      Method #1

      1. Let $u = x^{2}$.

         Then let $du = 2 x dx$ and substitute $\frac{du}{2}$:

         $$\int \frac{u^{2}}{4}\, du$$

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

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

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

               $$\int u^{2}\, du = \frac{u^{3}}{3}$$

            So, the result is: $\frac{u^{3}}{6}$

         Now substitute $u$ back in:

         $$\frac{x^{6}}{6}$$

      Method #2

      1. Let $u = x^{3}$.

         Then let $du = 3 x^{2} dx$ and substitute $\frac{du}{3}$:

         $$\int \frac{u}{9}\, du$$

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

            $$\int \frac{u}{3}\, du = \frac{\int u\, du}{3}$$

            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: $\frac{u^{2}}{6}$

         Now substitute $u$ back in:

         $$\frac{x^{6}}{6}$$

   So, the result is: $15 x^{6}$

2. Add the constant of integration:

   $$15 x^{6}+ \mathrm{constant}$$

The answer is:

$$15 x^{6}+ \mathrm{constant}$$