Integral of ln(y^2+1) dx

1. Use integration by parts:

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

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

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

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

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

      $$\int 1\, dy = y$$

   Now evaluate the sub-integral.

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

   $$\int \frac{2 y^{2}}{y^{2} + 1}\, dy = 2 \int \frac{y^{2}}{y^{2} + 1}\, dy$$

   1. Rewrite the integrand:

      $$\frac{y^{2}}{y^{2} + 1} = 1 - \frac{1}{y^{2} + 1}$$

   2. Integrate term-by-term:

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

         $$\int 1\, dy = y$$

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

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

         PiecewiseRule(subfunctions=[(ArctanRule(a=1, b=1, c=1, context=1/(y**2 + 1), symbol=y), True), (ArccothRule(a=1, b=1, c=1, context=1/(y**2 + 1), symbol=y), False), (ArctanhRule(a=1, b=1, c=1, context=1/(y**2 + 1), symbol=y), False)], context=1/(y**2 + 1), symbol=y)

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

      The result is: $y - \operatorname{atan}{\left(y \right)}$

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

3. Now simplify:

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

4. Add the constant of integration:

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

The answer is:

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