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?

It is sometimes better to take $x+a$ (for some constant $a$) as the antiderivative of $1$. In this case
$$
\int \arctan\sqrt{x+2}\,dx=(x+a)\arctan\sqrt{x+2}-\frac{1}{2}\int(x+a)\frac{1}{1+x+2}\cdot\frac{1}{\sqrt{x+2}}\,dx,
$$
so $a=3$ seems to be a good choice. 
I think you can continue from here.
