Ολοκλήρωμα

Tolaso J Kos · #1

Να υπολογιστεί το ολοκλήρωμα:

$\displaystyle{\mathcal{J} = \int_0^2 \frac{\ln (1+x)}{x^2-x+1} \, \mathrm{d}x}$

Tolaso J Kos · #2

Tolaso J Kos έγραψε: Σάβ Φεβ 15, 2020 8:12 pm Να υπολογιστεί το ολοκλήρωμα:

$\displaystyle{\mathcal{J} = \int_0^2 \frac{\ln (1+x)}{x^2-x+1} \, \mathrm{d}x}$

Επαναφορά!

neutonas · #3

$\displaystyle J = \int_0^2 \frac{\ln(x+1)}{x^2 - x + 1} \, dx = \int_0^2 \frac{\ln(x+1)}{(x - \frac{1}{2})^2 + \frac{3}{4}} \, dx$

$\displaystyle u = \frac{2x - 1}{\sqrt{3}} \Rightarrow x = \frac{u \sqrt{3} + 1}{2} \Rightarrow dx = \frac{\sqrt{3}}{2} \, du \quad x = 0 \Rightarrow u = -\frac{1}{\sqrt{3}}, \quad x = 2 \Rightarrow u = \sqrt{3} \quad \text{Thus:}$

$\displaystyle J = \int_{-\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{\ln \left( \frac{u \sqrt{3} + 1}{2} + 1 \right)}{\frac{3}{4} u^2 + \frac{3}{4}} \frac{\sqrt{3}}{2} \, du = \frac{2}{\sqrt{3}} \int_{-\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{\ln \left( \frac{u \sqrt{3} + 3}{2} \right)}{u^2 + 1} \, du = \frac{2}{\sqrt{3}} \int_{-\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{\ln (u \sqrt{3} + 3) - \ln 2}{u^2 + 1} \, du$

$\displaystyle = \frac{2}{\sqrt{3}} \int_{-\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{\ln (u \sqrt{3} + 3)}{u^2 + 1} \, du - \frac{2}{\sqrt{3}} \ln 2 \int_{-\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{1}{u^2 + 1} \, du = I_1 - I_2 = J$

$\displaystyle I_2 = \frac{2} {\sqrt{3}}\ln 2 \int_{-\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{1}{u^2 + 1} \, du = \frac{2} {\sqrt{3}} \ln 2 \, \arctan(u) \Bigg|_{-\frac{1}{\sqrt{3}}}^{\sqrt{3}} = \frac{2} {\sqrt{3}} \ln 2 \left( \frac{\pi}{3} - \left( -\frac{\pi}{6} \right) \right)$
$\displaystyle = \frac{2} {\sqrt{3}} \ln 2 \frac{\pi}{2} =\frac{\pi} {\sqrt{3}} \ln 2$

$\displaystyle I_1 = \frac{2}{\sqrt{3}} \int_{-\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{\ln (u \sqrt{3} + 3)}{u^2 + 1} \, du$
$\displaystyle \quad u = \tan \theta \quad du = \sec^2 \theta \, d\theta \quad u = -\frac{1}{\sqrt{3}} \Rightarrow \theta = -\frac{\pi}{6} \quad u = \sqrt{3} \Rightarrow \theta = \frac{\pi}{3}$

$\displaystyle I_1=\frac{2}{\sqrt{3}} \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\ln(\sqrt{3}\tan \theta + 3)}{\tan^2 \theta + 1} \sec^2 \theta \, d\theta = \frac{2}{\sqrt{3}} \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \ln(\sqrt{3}\tan \theta + 3) \, d\theta$

$\displaystyle = \frac{2}{\sqrt{3}} \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \ln \left( \frac{\sqrt{3} \sin \theta + 3 \cos \theta}{\cos \theta} \right) d\theta$

$\displaystyle = \frac{2}{\sqrt{3}} \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \ln \left( \sqrt{3} \sin \theta + 3 \cos \theta \right) d\theta - \frac{2}{\sqrt{3}} \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \ln (\cos \theta) d\theta$

$\displaystyle \text{But} \quad \sqrt{3} \sin \theta + 3 \cos \theta = A \cos (\theta + a) \quad \text{with}$

$\displaystyle A^2 = (\sqrt{3})^2 + 3^2 = 12 \Rightarrow A = 2 \sqrt{3} \quad \text{and} \quad \tan a = -\frac{\sqrt{3}}{3} \Rightarrow a = -\frac{\pi}{6}$

$\displaystyle \text{Thus :} \quad I_1 = \frac{2}{\sqrt{3}} \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \ln \left( 2\sqrt{3} \cos (\theta - \frac{\pi}{6}) \right) d\theta - \frac{2}{\sqrt{3}} \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \ln (\cos \theta) d\theta$

$\displaystyle = \frac{2}{\sqrt{3}} \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \ln (2\sqrt{3}) d\theta + \frac{2}{\sqrt{3}} \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \ln (\cos (\theta - \frac{\pi}{6})) d\theta - \frac{2}{\sqrt{3}} \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \ln (\cos \theta) d\theta$
$\displaystyle \text{But} \quad \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \ln (\cos (\theta - \frac{\pi}{6})) d\theta = \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \ln (\cos ( \frac{\pi}{6}-\theta ) ) d\theta = - \int_{\frac{\pi}{3}}^{-\frac{\pi}{6}} \ln (\cos u ) du= \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \ln (\cos u ) du$
$\displaystyle \text{With :} \quad u = \frac{\pi}{6}-\theta \Rightarrow du = - d\theta \quad \theta = -\frac{\pi}{6} \Rightarrow u = \frac{\pi}{3}, \quad \theta=\frac{\pi}{3} \Rightarrow u = -\frac{\pi}{6}$
$\displaystyle \text{Thus :} \quad I_1 = \frac{2}{\sqrt{3}} \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \ln (2\sqrt{3}) d\theta = \frac{2}{\sqrt{3}} \ln (2\sqrt{3}) \frac{3 \pi}{6} = \frac{\pi \ln (2\sqrt{3})}{\sqrt{3}}$

$\displaystyle \text{So :} \quad J= I_1 - I_2 = \frac{\pi \ln (2\sqrt{3})}{\sqrt{3}} - \frac{\pi \ln (2)}{\sqrt{3}} = \frac{\pi \ln (\sqrt{3})}{\sqrt{3}} \Rightarrow \, \boxed{J=\frac{\pi \ln (\sqrt{3})}{\sqrt{3}} }$

#4

neutonas έγραψε: Πέμ Ιούλ 08, 2021 11:00 pm Παραθέτω μία λύση ως συνημμένο.

Ας υπενθυμίσω ότι οι κανονισμοί μας (πολύ σωστά) δεν επιτρέπουν συνημμένο χειρόγραφο. Παράκληση, λοιπόν, να το διευθετήσεις (Latex) γιατί βλέπω οι Γενικοί Συντονιστές να σβήνουν το μήνυμά σου.

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