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 $\frac{dx}{a^2+x^2}$
 A: Hint: $$x^2\pm\sqrt{2}x+1 = (x\pm 1/\sqrt{2})^2+1/2$$
details:
$$\int_{-1}^1 \frac {dx}{x^2+\sqrt{2}x+1}
= \int_{-1}^1 \frac {dx}{(x+ 1/\sqrt{2})^2+1/2}
= \int_{-1+1/\sqrt{2}}^{1+1/\sqrt{2}}
 \frac {du}{u^2+1/2}
\\
 = \left[1/\sqrt{2} \arctan \sqrt{2} u
\right]_{-1+1/\sqrt{2}}^{1+1/\sqrt{2}}
$$
A: Well, it isn't yet, but you can change it into the proper form. In fact, you can change any square polynomial $x^2+2bx + c$ into $(x+b)^2 + (c-b^2)$ which equals $(x+b)^2 + \sqrt{c-b^2}^2$. Now, just introduce  new variable $t=x+b^2$.
A: Do you remember how to complete the square? $$x^2 + ax + b = x^2 + ax + (\frac{a}{2})^2 - (\frac{a}{2})^2 + b = (x + \frac{a}{2})^2 + b - (\frac{a}{2})^2$$
Do the same for this question, and then you should see an easy chance to substitute and the integration will pop out of the page!
A: 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 $$\frac{1}{2} \int_{-1}^1 \frac{dx}{1+\sqrt{2}x+x^2} =\frac{\pi }{2 \sqrt{2}}$$
