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Informaţii, definiţii, teoreme, formule, exerciţii şi probleme rezolvate din matematica de liceu.

Data publicarii: 28 Noiembrie, 2015

EXERCITIUL 28

Suport teoretic:

Integrale definite,functii gradul 2,logaritmi naturali,functia arctg.

Enunt: 

Sa se rezolve in multimea numerelor reale ecuatia:

\int_0^x{\frac{t+1}{t^2+t+1}}dt=ln{\sqrt{x^2+x+1}}\cdot\int_0^x{\frac{t+1}{t^2+t+1}}dt=ln{\sqrt{x^2+x+1}}\cdot

Raspuns: 

S = {0}. 

Rezolvare:

I=\int_0^x{\frac{t+1}{t^2+t+1}}dt=\frac{1}{2}\big(\int_0^x{\frac{2t+1}{t^2+t+1}}dt+\int_0^x{\frac{1}{t^2+t+1}}dt\big)=\cdots=I=\int_0^x{\frac{t+1}{t^2+t+1}}dt=\frac{1}{2}\big(\int_0^x{\frac{2t+1}{t^2+t+1}}dt+\int_0^x{\frac{1}{t^2+t+1}}dt\big)=\cdots=

=ln{\sqrt{t^2+t+1}}\big|_0^x+{\frac{1}{2}}\cdot{\int_0^x{\frac{1}{(t+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}}dt}==ln{\sqrt{t^2+t+1}}\big|_0^x+{\frac{1}{2}}\cdot{\int_0^x{\frac{1}{(t+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}}dt}= ln{\sqrt{x^2+x+1}}+{\frac{\sqrt{3}}{3}}\cdot{arctg{\frac{2t+1}{\sqrt{3}}}}\big|_0^x=ln{\sqrt{x^2+x+1}}+{\frac{\sqrt{3}}{3}}\cdot{arctg{\frac{2t+1}{\sqrt{3}}}}\big|_0^x=

=\cdots=ln{\sqrt{x^2+x+1}}+{\frac{\sqrt{3}}{3}}\cdot{arctg{\frac{2x+1}{\sqrt{3}}}-\frac{\sqrt{3}\pi}{18}}\cdot=\cdots=ln{\sqrt{x^2+x+1}}+{\frac{\sqrt{3}}{3}}\cdot{arctg{\frac{2x+1}{\sqrt{3}}}-\frac{\sqrt{3}\pi}{18}}\cdot

Rezulta echivalenta:

I=ln{\sqrt{x^2+x+1}}\Leftrightarrow{ln{\sqrt{x^2+x+1}}+{\frac{\sqrt{3}}{3}}\cdot{arctg{\frac{2x+1}{\sqrt{3}}}-\frac{\sqrt{3}\pi}{18}}}=ln{\sqrt{x^2+x+1}}\cdotI=ln{\sqrt{x^2+x+1}}\Leftrightarrow{ln{\sqrt{x^2+x+1}}+{\frac{\sqrt{3}}{3}}\cdot{arctg{\frac{2x+1}{\sqrt{3}}}-\frac{\sqrt{3}\pi}{18}}}=ln{\sqrt{x^2+x+1}}\cdot

Un calcul simplu conduce imediat la solutia x = 0. 


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