Integral of (arccossqrtx)/sqrt(x+1) dx

1. Let $u = \sqrt{x}$.

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

   $$\int \frac{2 u \operatorname{acos}{\left(u \right)}}{\sqrt{u^{2} + 1}}\, du$$

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

      $$\int \frac{u \operatorname{acos}{\left(u \right)}}{\sqrt{u^{2} + 1}}\, du = 2 \int \frac{u \operatorname{acos}{\left(u \right)}}{\sqrt{u^{2} + 1}}\, du$$

      1. Use integration by parts:

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

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

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

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

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

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

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

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

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

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

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

               So, the result is: $\sqrt{u}$

            Now substitute $u$ back in:

            $$\sqrt{u^{2} + 1}$$

         Now evaluate the sub-integral.

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

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

         TrigSubstitutionRule(theta=_theta, func=sin(_theta), rewritten=sqrt(sin(_theta)**2 + 1), substep=EllipticERule(a=1, d=-1, context=sqrt(sin(_theta)**2 + 1), symbol=_theta), restriction=(_u > -1) & (_u < 1), context=sqrt(_u**2 + 1)/sqrt(1 - _u**2), symbol=_u)

         So, the result is: $- \begin{cases} E\left(\operatorname{asin}{\left(u \right)}\middle| -1\right) & \text{for}\: u > -1 \wedge u < 1 \end{cases}$

      So, the result is: $2 \sqrt{u^{2} + 1} \operatorname{acos}{\left(u \right)} + 2 \left(\begin{cases} E\left(\operatorname{asin}{\left(u \right)}\middle| -1\right) & \text{for}\: u > -1 \wedge u < 1 \end{cases}\right)$

   Now substitute $u$ back in:

   $$2 \sqrt{x + 1} \operatorname{acos}{\left(\sqrt{x} \right)} + 2 \left(\begin{cases} E\left(\operatorname{asin}{\left(\sqrt{x} \right)}\middle| -1\right) & \text{for}\: x \geq 0 \wedge x < 1 \end{cases}\right)$$

2. Now simplify:

   $$\begin{cases} 2 \left(\sqrt{x + 1} \operatorname{acos}{\left(\sqrt{x} \right)} + E\left(\operatorname{asin}{\left(\sqrt{x} \right)}\middle| -1\right)\right) & \text{for}\: x \geq 0 \wedge x < 1 \end{cases}$$

3. Add the constant of integration:

   $$\begin{cases} 2 \left(\sqrt{x + 1} \operatorname{acos}{\left(\sqrt{x} \right)} + E\left(\operatorname{asin}{\left(\sqrt{x} \right)}\middle| -1\right)\right) & \text{for}\: x \geq 0 \wedge x < 1 \end{cases}+ \mathrm{constant}$$

The answer is:

$$\begin{cases} 2 \left(\sqrt{x + 1} \operatorname{acos}{\left(\sqrt{x} \right)} + E\left(\operatorname{asin}{\left(\sqrt{x} \right)}\middle| -1\right)\right) & \text{for}\: x \geq 0 \wedge x < 1 \end{cases}+ \mathrm{constant}$$