456* Integrals.

by Virgil Nicula, Jul 31, 2017, 1:31 PM

P1 (Kunihiko Chikaya). Calculate the definite integral $I=\int\limits_{\sqrt 2-1}^{\sqrt 2+1}f(x)\ \mathrm{dx}\ ,$ where $f(x)=\frac {x^4+x^2+2}{x^4+2x^2+1}\ .$

Proof. $f(x)=1+g(x)\ ,$ where $g(x)=\frac {1-x^2}{\left(x^2+1\right)^2}$ and $\left|x-\sqrt 2\right|\le 1\iff \boxed{I=2+J}\ (*)\ \ ,$ where $J=\int\limits_{\sqrt 2-1}^{\sqrt 2+1}g(x)\ \mathrm{dx}\ .$ Thus, $J=\int\limits_{\sqrt 2-1}^{\sqrt 2+1}\frac {1-x^2}{\left(x^2+1\right)^2}\ \mathrm{dx}=$

$\int\limits_{\sqrt 2-1}^{\sqrt 2+1}\frac {\frac 1{x^2}-1}{\left(x+\frac 1x\right)^2}\ \mathrm{dx}=$ $\int\limits_{\sqrt 2-1}^{\sqrt 2+1}\frac {-\left(x+\frac 1x\right)'}{\left(x+\frac 1x\right)^2}\ \mathrm{dx}\ \stackrel{t=\phi (x)=x+\frac 1x}{=}\ \int\limits_{2\sqrt 2}^{2\sqrt 2}\frac {-1}{t^2}\ \mathrm{dt}=$ $\left\{\ \frac 1t\ \right|_{2\sqrt 2}^{2\sqrt 2}=0\implies \boxed{\ J=0\ }\ (1)\ .$ In conclusion, $J=0$ and $I=2+J\implies I=2\ .$
This post has been edited 20 times. Last edited by Virgil Nicula, Feb 8, 2018, 7:30 PM

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