Integral of x^3/(x^2+1) 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 $du$:

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

      1. Rewrite the integrand:

         $$\frac{u}{2 u + 2} = \frac{1}{2} - \frac{1}{2 \left(u + 1\right)}$$

      2. Integrate term-by-term:

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

            $$\int \frac{1}{2}\, du = \frac{u}{2}$$

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

            $$\int \left(- \frac{1}{2 \left(u + 1\right)}\right)\, du = - \frac{\int \frac{1}{u + 1}\, du}{2}$$

            1. Let $u = u + 1$.

               Then let $du = du$ 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:

               $$\log{\left(u + 1 \right)}$$

            So, the result is: $- \frac{\log{\left(u + 1 \right)}}{2}$

         The result is: $\frac{u}{2} - \frac{\log{\left(u + 1 \right)}}{2}$

      Now substitute $u$ back in:

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

   Method #2

   1. Rewrite the integrand:

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

   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 times a function is the constant times the integral of the function:

         $$\int \left(- \frac{x}{x^{2} + 1}\right)\, dx = - \int \frac{x}{x^{2} + 1}\, dx$$

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

            $$\int \frac{x}{x^{2} + 1}\, dx = \frac{\int \frac{2 x}{x^{2} + 1}\, dx}{2}$$

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

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

               $$\int \frac{1}{2 u}\, du$$

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

               Now substitute $u$ back in:

               $$\log{\left(x^{2} + 1 \right)}$$

            So, the result is: $\frac{\log{\left(x^{2} + 1 \right)}}{2}$

         So, the result is: $- \frac{\log{\left(x^{2} + 1 \right)}}{2}$

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

2. Add the constant of integration:

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

The answer is:

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