<s>[INST] Use the integration formula $\frac{1}{a}\arctan\frac{x}{a}$ to solve $\frac{1}{2} \int_{-1}^1 \mathrm{ \frac{dx}{1+\sqrt{2}x+x^2} }\, $ As question states, I am trying to figure out how to use the integration formula to solve the integral. My issue is that the integral isn't of the form $\frac{dx}{a^2+x^2}$
 [/INST] If I amy interfer in this discussion, you have been shown by various answers and comments that through the appropriate change of variable dictated by completing the square
$$\frac{1}{2} \int_{}  \frac{dx}{1+\sqrt{2}x+x^2} =\frac{\tan ^{-1}\left(x\sqrt{2} +1\right)}{\sqrt{2}}$$ So, putting the bounds $$\frac{1}{2} \int_{-1}^1 \frac{dx}{1+\sqrt{2}x+x^2} =\frac{1}{\sqrt{2}}[\tan^{-1}(1+\sqrt{2})-\tan^{-1}(1-\sqrt{2})]$$ Now, use the fact that $$\tan^{-1}(a)-\tan^{-1}(b)=\tan ^{-1}\left(\frac{a-b}{1+a \times b}\right)$$ Put $a=1+\sqrt{2}$, $b=1-\sqrt{2}$ and you will find that $\frac{a-b}{1+a\times b}$ is infinity and for such an argument, the $\tan ^{-1}$ is equal to $\frac{\pi}{2}$. You then end with $$\frac{1}{2} \int_{-1}^1 \frac{dx}{1+\sqrt{2}x+x^2} =\frac{\pi }{2 \sqrt{2}}$$
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