$\def\d{\mathrm{d}}$Any hint on how to compute this integration?$$\int \arcsin\left(\frac{x+2}{x^2+4x+13}\right)\,\d x$$
This is what I have done:$$ \int \arcsin\left(\frac{x+2}{x^2+4x+13}\right)\,\d x=\int \arcsin\left(\frac{x+2}{(x+2)^2+9}\right)\,\d x. $$ Substitute $x+2=3 \tan\theta$ and $\d x=3\sec^2\theta\,\d\theta$,\begin{align*} &\mathrel{\phantom{=}} \int \arcsin\left(\frac{x+2}{(x+2)^2+9}\right)\,\d x\\ &=3\int \arcsin\left(\frac{3\tan\theta}{9\tan^2\theta+9}\right)\sec^2\theta\,\d\theta\\ &=3\int \arcsin\left(\frac{3\tan\theta}{9\sec^2\theta}\right)\sec^2\theta\,\d\theta\\ &=3\int \arcsin\left(\frac{\sin\theta\cos\theta}{3}\right)\sec^2\theta\,\d\theta\\ &=3\int \arcsin\left(\frac{\sin2\theta}{6}\right)\sec^2\theta\,\d\theta. \end{align*}
What can I do from here?