Integral of xln(1+x²) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \log{\left(x^{2} + 1 \right)}$.

      Then let $du = \frac{2 x dx}{x^{2} + 1}$ and substitute $\frac{du}{2}$:

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

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

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

         1. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

            Then $\operatorname{du}{\left(u \right)} = 1$.

            To find $v{\left(u \right)}$:

            1. The integral of the exponential function is itself.

               $$\int e^{u}\, du = e^{u}$$

            Now evaluate the sub-integral.

         2. The integral of the exponential function is itself.

            $$\int e^{u}\, du = e^{u}$$

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

      Now substitute $u$ back in:

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

   Method #2

   1. Use integration by parts:

      $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

      Let $u{\left(x \right)} = \log{\left(x^{2} + 1 \right)}$ and let $\operatorname{dv}{\left(x \right)} = x$.

      Then $\operatorname{du}{\left(x \right)} = \frac{2 x}{x^{2} + 1}$.

      To find $v{\left(x \right)}$:

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

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

      Now evaluate the sub-integral.

   2. 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:

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

2. Add the constant of integration:

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

The answer is:

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