Ολοκλήρωμα

mick7 · #1

Να υπολογιστεί το

$\displaystyle \int \frac{\sqrt{x^2+2}}{x^2+1}dx$

Tolaso J Kos · #2

mick7 έγραψε: ↑
Δευ Ιουν 07, 2021 3:29 pm
Να υπολογιστεί το

$\displaystyle \int \frac{\sqrt{x^2+2}}{x^2+1}dx$

Έχομεν και λέμε:

$\displaystyle{\begin{aligned} \int \frac{\sqrt{x^2+2}}{x^2+1} \, \mathrm{d}x &= \int \frac{\sqrt{x^2+2} \sqrt{x^2+2}}{\sqrt{x^2+2} \left ( 1+x^2 \right ) } \, \mathrm{d}x \\ &= \int \frac{x^2+2}{\left ( 1+x^2 \right ) \sqrt{x^2+2}} \, \mathrm{d}x \\ &= \int \frac{x^2+1+1}{\left ( 1+x^2 \right ) \sqrt{x^2+2}} \, \mathrm{d}x \\ &= \int \left ( \frac{1}{\sqrt{x^2+2}} + \frac{1}{\left ( 1+x^2 \right ) \sqrt{x^2+2}} \right ) \, \mathrm{d}x \\ &= \sinh^{-1} \frac{x}{\sqrt{2}} + \int \frac{\mathrm{d}x}{\left ( 1+x^2 \right ) \sqrt{x^2+2}} \end{aligned}}$
Για το τελευταίο ολοκλήρωμα έχουμε διαδοχικά:

$\displaystyle{\begin{aligned} \int \frac{\mathrm{d}x}{\left ( 1+x^2 \right ) \sqrt{x^2+2}} &\overset{x=\sqrt{2} \tan \theta}{=\! =\! =\! =\! =\! =\! =\! =\!} \int \frac{\sqrt{2} \sec^2 \theta}{\left ( \tan^2 \theta + \sec^2 \theta \right ) \sqrt{2 \tan^2 \theta + 2}} \, \mathrm{d}\theta \\ &= \int \frac{\sec^2 \theta}{\left ( \tan^2 \theta + \sec^2 \theta \right ) \sqrt{\tan^2 \theta + 1}} \, \mathrm{d}\theta \\ &= \int \frac{\sec^2 \theta}{\left ( \tan^2 \theta + \sec^2 \theta \right ) \sec \theta} \, \mathrm{d}\theta \\ &= \int \frac{\sec \theta}{\tan^2 \theta + \sec^2 \theta} \, \mathrm{d} \theta \\ &= \int \frac{\sec \theta}{\tan^2 \theta + \sec^2 \theta}\cdot \left ( \frac{\cos^2 \theta}{\cos^2 \theta} \right ) \, \mathrm{d}\theta \\ &= \int \frac{\cos \theta}{\sin^2 \theta + 1} \, \mathrm{d} \theta \\ &= \arctan \sin \theta + c \end{aligned}}$
Όμως,

$\displaystyle{x = \sqrt 2\tan\theta \Rightarrow \tan\theta = \frac x{\sqrt 2} \Rightarrow \sin \theta = \frac{x}{\sqrt{x^2 + 2}} }$
Τελικά,

$\displaystyle{\int \frac{\sqrt{x^2+2}}{x^2+1} \, \mathrm{d}x = \sinh^{-1} \frac{x}{\sqrt{2}} + \arctan \frac{x}{\sqrt{x^2 + 2}} + c \; , \; c \in \mathbb{R}}$

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