Integral of ln(4x^2+1)dx dx

1. Use integration by parts:

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

   Let $u{\left(x \right)} = \log{\left(4 x^{2} + 1 \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$.

   Then $\operatorname{du}{\left(x \right)} = \frac{8 x}{4 x^{2} + 1}$.

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

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

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

   Now evaluate the sub-integral.

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

   $$\int \frac{8 x^{2}}{4 x^{2} + 1}\, dx = 8 \int \frac{x^{2}}{4 x^{2} + 1}\, dx$$

   1. Rewrite the integrand:

      $$\frac{x^{2}}{4 x^{2} + 1} = \frac{1}{4} - \frac{1}{4 \cdot \left(4 x^{2} + 1\right)}$$

   2. Integrate term-by-term:

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

         $$\int \frac{1}{4}\, dx = \frac{x}{4}$$

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

         $$\int \left(- \frac{1}{4 \cdot \left(4 x^{2} + 1\right)}\right)\, dx = - \frac{\int \frac{1}{4 x^{2} + 1}\, dx}{4}$$

         1. The integral of $\frac{1}{x^{2} + 1}$ is $\frac{\operatorname{atan}{\left(2 x \right)}}{2}$.

         So, the result is: $- \frac{\operatorname{atan}{\left(2 x \right)}}{8}$

      The result is: $\frac{x}{4} - \frac{\operatorname{atan}{\left(2 x \right)}}{8}$

   So, the result is: $2 x - \operatorname{atan}{\left(2 x \right)}$

3. Now simplify:

   $$x \log{\left(4 x^{2} + 1 \right)} - 2 x + \operatorname{atan}{\left(2 x \right)}$$

4. Add the constant of integration:

   $$x \log{\left(4 x^{2} + 1 \right)} - 2 x + \operatorname{atan}{\left(2 x \right)}+ \mathrm{constant}$$

The answer is:

$$x \log{\left(4 x^{2} + 1 \right)} - 2 x + \operatorname{atan}{\left(2 x \right)}+ \mathrm{constant}$$