Calcul de integrala 2
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Calcul de integrala 2
Sa se calculeze \( I=\int \frac{\arctan x-\frac{x}{x^2+1}}{(x+\arctan x)(x-\arctan x)}dx \) pentru \( x\in(0,\infty) \)
Last edited by liv on Wed Jan 09, 2008 7:51 pm, edited 2 times in total.
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Definim functiile \( f, g: (0,\infty)\to\mathbb{R} \) definite prin:
\( f(x)=x+\arctan x \) si \( g(x)=x-\arctan x \). Este limpede ca \( f,g \) sunt derivabile (functii elementare). Astfel, avem
\( f^ \prime (x)=1+\frac{1}{x^{2}+1} \) si \( g^ \prime (x)=1-\frac{1}{x^{2}+1} \).
Se observa ca
\( f^\prime(x)g(x)-g^\prime(x)f(x)= \)
\( \left(x-\arctan x+\frac{x}{x^{2}+1}-\frac{\arctan x}{x^{2}+1}\right)-\left(x+\arctan x-\frac{x}{x^{2}+1}-\frac{\arctan x}{x^{2}+1}\right)= \)
\( \frac{2x}{x^{2}+1}-2\arctan x, \forall x>0 \).
Astfel, integrala noastra este egala, de fapt, cu
\( I=\frac{1}{2}\int\frac{g^ \prime(x)f(x)-f^ \prime(x)g(x)}{f(x)g(x)}dx=\frac{1}{2}\int\left(\frac{g(x)}{g^ \prime(x)}-\frac{f(x)}{f^ \prime(x)}\right)dx=ln\frac{g(x)}{f(x)}+\mathcal{C} \).
\( f(x)=x+\arctan x \) si \( g(x)=x-\arctan x \). Este limpede ca \( f,g \) sunt derivabile (functii elementare). Astfel, avem
\( f^ \prime (x)=1+\frac{1}{x^{2}+1} \) si \( g^ \prime (x)=1-\frac{1}{x^{2}+1} \).
Se observa ca
\( f^\prime(x)g(x)-g^\prime(x)f(x)= \)
\( \left(x-\arctan x+\frac{x}{x^{2}+1}-\frac{\arctan x}{x^{2}+1}\right)-\left(x+\arctan x-\frac{x}{x^{2}+1}-\frac{\arctan x}{x^{2}+1}\right)= \)
\( \frac{2x}{x^{2}+1}-2\arctan x, \forall x>0 \).
Astfel, integrala noastra este egala, de fapt, cu
\( I=\frac{1}{2}\int\frac{g^ \prime(x)f(x)-f^ \prime(x)g(x)}{f(x)g(x)}dx=\frac{1}{2}\int\left(\frac{g(x)}{g^ \prime(x)}-\frac{f(x)}{f^ \prime(x)}\right)dx=ln\frac{g(x)}{f(x)}+\mathcal{C} \).
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.