<s>[INST] Integrate $\int\frac{dx}{(x^2+1)\sqrt{x^2+2}}$ I would like some guidance regarding the following integral:
$$\int\frac{dx}{(x^2+1)\sqrt{x^2+2}}$$
EDIT: The upper problem was derived from the following integral $$\int\frac{\sqrt{x^2+2}}{x^2+1}dx$$
Where I rationalized the numerator which followed into: $$\int\frac{dx}{\sqrt{x^2+2}}+\int\frac{dx}{(x^2+1)\sqrt{x^2+2}}$$
 [/INST] If you aren't aware of the posted formula for calculating an integral of this form (which would completely understandable!), we can start with more familiar approaches, such as using trigonometric substitution.
Try putting $x = \sqrt 2\tan \theta$. Then $$dx = \sqrt 2 \sec^2 \theta\,d\theta$$
$$\text{ and }\,x^2 + 1 = 2\tan^2 \theta + 1 = \tan^2 \theta + \tan^2 \theta + 1 = \tan^2 \theta + \sec^2 \theta$$
$$\begin{align}\int \frac{dx}{(x^2 + 1)\sqrt {x^2 + 2}} &=\int \frac{\sqrt 2\sec^2 \theta\,d\theta}{(\tan^2 \theta+\sec^2 \theta)\sqrt{2\tan^2 \theta + 2}}\\ \\ 
&= \int \frac{\sec^2 \theta\,d\theta}{(\tan^2 \theta+\sec^2 \theta)\underbrace{\sqrt{\tan^2 \theta + 1}}_{\large = \,\sqrt{\sec^2 x}}}\\ \\ &=\int \frac{ \sec^2 \theta\,d\theta}{(\tan^2 \theta + \sec^2 \theta)\sec\theta} \\ \\ 
&= \int \dfrac{\sec \theta\,d\theta}{\tan^2 \theta + \sec^2 \theta}\cdot\left(\frac{\cos^2 \theta}{\cos^2 \theta}\right) \\ \\ &= \int\frac{\cos \theta\,d\theta}{\sin^2\theta + 1} \\ \\
&= \int\frac{d(\sin \theta)}{(\sin \theta)^2 + 1}\\ \\ &= \tan^{-1}(\sin \theta) + C \end{align}$$
Now, in terms of $x$, we have $$x = \sqrt 2\tan\theta \implies \tan\theta = \dfrac x{\sqrt 2} \implies \sin \theta = \frac{x}{\sqrt{x^2 + 2}}$$
That gives us the final answer in terms of $x$: $$\tan^{-1}(\sin \theta) + C = \tan^{-1}\left(\frac{x}{\sqrt{x^2 + 2}}\right) + C$$
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