<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] Hint: $$x^2\pm\sqrt{2}x+1 = (x\pm 1/\sqrt{2})^2+1/2$$
details:
$$\int_{-1}^1 \frac {dx}{x^2+\sqrt{2}x+1}
= \int_{-1}^1 \frac {dx}{(x+ 1/\sqrt{2})^2+1/2}
= \int_{-1+1/\sqrt{2}}^{1+1/\sqrt{2}}
 \frac {du}{u^2+1/2}
\\
 = \left[1/\sqrt{2} \arctan \sqrt{2} u
\right]_{-1+1/\sqrt{2}}^{1+1/\sqrt{2}}
$$
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