Integral of x*arccos(x) dx

1. Use integration by parts:

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

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

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

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

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

   SqrtQuadraticDenomRule(a=1, b=0, c=-1, coeffs=[1, 0, 0], context=x**2/sqrt(1 - x**2), symbol=x)

   So, the result is: $\frac{x \sqrt{1 - x^{2}}}{4} - \frac{\operatorname{asin}{\left(x \right)}}{4}$

3. Add the constant of integration:

   $$\frac{x^{2} \operatorname{acos}{\left(x \right)}}{2} - \frac{x \sqrt{1 - x^{2}}}{4} + \frac{\operatorname{asin}{\left(x \right)}}{4}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{2} \operatorname{acos}{\left(x \right)}}{2} - \frac{x \sqrt{1 - x^{2}}}{4} + \frac{\operatorname{asin}{\left(x \right)}}{4}+ \mathrm{constant}$$