Integral of x^2/2x^3+1dx dx

1. Integrate term-by-term:

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

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

      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: $\frac{x^{6}}{12}$

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

      $$\int 1 \cdot 1\, dx = x$$

   The result is: $\frac{x^{6}}{12} + x$

2. Add the constant of integration:

   $$\frac{x^{6}}{12} + x+ \mathrm{constant}$$

The answer is:

$$\frac{x^{6}}{12} + x+ \mathrm{constant}$$