Integral of x^2/2x-1 dx

1. Integrate term-by-term:

   1. There are multiple ways to do this integral.

      Method #1

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

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

         $$\int u\, du$$

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

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

         Now substitute $u$ back in:

         $$\frac{x^{4}}{8}$$

      Method #2

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

         Then let $du = 2 x dx$ and substitute $\frac{du}{4}$:

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

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

            $$\int u\, du = \frac{\int u\, du}{4}$$

            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: $\frac{u^{2}}{8}$

         Now substitute $u$ back in:

         $$\frac{x^{4}}{8}$$

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

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

   The result is: $\frac{x^{4}}{8} - x$

2. Add the constant of integration:

   $$\frac{x^{4}}{8} - x+ \mathrm{constant}$$

The answer is:

$$\frac{x^{4}}{8} - x+ \mathrm{constant}$$