Solve definite integral: $\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$ I need to solve:
$$\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$$
Here is my steps, first of all consider just the indefinite integral:
$$\int \arctan(\sqrt{x+2})dx = \int \arctan(\sqrt{x+2}) \cdot 1\ dx$$
$$f(x) = \arctan(\sqrt{x+2})$$
$$f'(x) = \frac{1}{1+x+2} \cdot \frac{1}{2\sqrt{x+2}} = \frac{1}{(2x+6)\sqrt{x+2}}$$
$$g'(x) = 1$$
$$g(x) = x$$
So:
$$\bigg[\arctan(\sqrt{x+2}) \cdot x\bigg]_{-1}^{1} - \int_{-1}^{1} \frac{x}{(2x-6)\sqrt{x+2}}\ dx$$
How should I proceed with the new integral?

Integration by parts yields
\begin{equation}
\int_{-1}^{1}\arctan\sqrt{x+2}\ dx=x\arctan\sqrt{x+2}\ \bigg|_{-1}^{1}-\frac12\int_{-1}^{1}\frac{x}{(x-1)\sqrt{x+2}}\ dx
\end{equation}
Making substitution $u^2=x+2$, then
\begin{align}
\int_{-1}^{1}\arctan\sqrt{x+2}\ dx&=\frac{\pi}{3}+\frac{\pi}{4}-\int_{1}^{\sqrt3}\frac{u^2-2}{u^2-3}\ du\\[10pt]
&=\frac{7\pi}{12}-\int_{1}^{\sqrt3}\frac{u^2-3+1}{u^2-3}\ du\\[10pt]
&=\frac{7\pi}{12}-\int_{1}^{\sqrt3}\ du-\int_{1}^{\sqrt3}\frac{1}{u^2-3}\ du\\[10pt]
\end{align}
Can you take it from here?
