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Data publicarii: 23 Martie, 2012

EXERCITIUL 4

Suport teoretic:

Schimbari variabila,integrare parti,integrale nedefinite.

Enunt:

Sa se calculeze pe intervalul (0,+oo) integrala:

I=\int{\sqrt{x+\sqrt{x+1}}}{dx}.I=\int{\sqrt{x+\sqrt{x+1}}}{dx}.

Raspuns:

I={\frac{2}{3}}\cdot{(x+\sqrt{x+1})}^{\frac{3}{2}}I={\frac{2}{3}}\cdot{(x+\sqrt{x+1})}^{\frac{3}{2}} +{\frac{1}{4}}\cdot{(2\sqrt{x+1}+1)}+{\frac{1}{4}}\cdot{(2\sqrt{x+1}+1)} \cdot{\sqrt{x+\sqrt{x+1}}}-\cdot{\sqrt{x+\sqrt{x+1}}}- {\frac{5}{4}}\cdot{{ln}{\frac{{2}\sqrt{x+1}+1-\sqrt{5}}{{2}\sqrt{x+1}+1+\sqrt{5}}}}+\mathcal{C}.{\frac{5}{4}}\cdot{{ln}{\frac{{2}\sqrt{x+1}+1-\sqrt{5}}{{2}\sqrt{x+1}+1+\sqrt{5}}}}+\mathcal{C}.

Rezolvare:

Cu schimbarea de variabila

\sqrt{x+1}=t\ge{1},\sqrt{x+1}=t\ge{1},

obtinem x = t² - 1, dx = 2tdt si, de aici, succesiv:

I_1=\int{2t\sqrt{t^2+t-1}}dt=\int{(2t+1-1)\sqrt{t^2+t-1}}dt=I_1=\int{2t\sqrt{t^2+t-1}}dt=\int{(2t+1-1)\sqrt{t^2+t-1}}dt=

=\int{(2t+1)\sqrt{t^2+t-1}}dt-\int{\sqrt{t^2+t-1}}dt==\int{(2t+1)\sqrt{t^2+t-1}}dt-\int{\sqrt{t^2+t-1}}dt=

=\int{(t^2+t-1)}^{\frac{1}{2}}d(t^2+t-1)+\int{\sqrt{t^2+t-1}}dt==\int{(t^2+t-1)}^{\frac{1}{2}}d(t^2+t-1)+\int{\sqrt{t^2+t-1}}dt=

={\frac{2}{3}}\cdot{(t^2+t-1)}^{\frac{3}{2}}+\int{\sqrt{t^2+t-1}}dt={\frac{2}{3}}\cdot{(t^2+t-1)}^{\frac{3}{2}}+J.\;(*)={\frac{2}{3}}\cdot{(t^2+t-1)}^{\frac{3}{2}}+\int{\sqrt{t^2+t-1}}dt={\frac{2}{3}}\cdot{(t^2+t-1)}^{\frac{3}{2}}+J.\;(*)

Calculam in continuare noua integrala obtinuta:

J=\int{\sqrt{t^2+t-1}}dt=\cdots=\int{\sqrt{(t+\frac{1}{2})^2-(\frac{\sqrt{5}}{2})^2}}dt.J=\int{\sqrt{t^2+t-1}}dt=\cdots=\int{\sqrt{(t+\frac{1}{2})^2-(\frac{\sqrt{5}}{2})^2}}dt.

Cu noua schimbare de variabila data de notatia

t+\frac{1}{2}=u\ge{\frac{3}{2}},t+\frac{1}{2}=u\ge{\frac{3}{2}},

obtinem, succesiv:

J_1=\int{\sqrt{u^2-(\frac{\sqrt{5}}{2})^2}}du=\cdots=\int{u\cdot{\frac{2u}{2\sqrt{u^2-(\frac{\sqrt{5}}{2})^2}}}}du-{\frac{5}{4}}\cdot{\int{\frac{1}{\sqrt{u^2-(\frac{\sqrt{5}}{2})^2}}}}du=\cdots=J_1=\int{\sqrt{u^2-(\frac{\sqrt{5}}{2})^2}}du=\cdots=\int{u\cdot{\frac{2u}{2\sqrt{u^2-(\frac{\sqrt{5}}{2})^2}}}}du-{\frac{5}{4}}\cdot{\int{\frac{1}{\sqrt{u^2-(\frac{\sqrt{5}}{2})^2}}}}du=\cdots=

=u\sqrt{u^2-\frac{5}{4}}-J_1-{\frac{\sqrt{5}}{4}}\cdot{ln|\frac{2u-\sqrt{5}}{2u+\sqrt{5}}|}.=u\sqrt{u^2-\frac{5}{4}}-J_1-{\frac{\sqrt{5}}{4}}\cdot{ln|\frac{2u-\sqrt{5}}{2u+\sqrt{5}}|}.

Rezulta de aici:

J_1=\cdots={\frac{1}{2}}\cdot{(u\sqrt{u^2-\frac{5}{4}}-{\frac{\sqrt{5}}{4}}\cdot{ln|\frac{2u-\sqrt{5}}{2u+\sqrt{5}}|})}+\mathcal{C}.J_1=\cdots={\frac{1}{2}}\cdot{(u\sqrt{u^2-\frac{5}{4}}-{\frac{\sqrt{5}}{4}}\cdot{ln|\frac{2u-\sqrt{5}}{2u+\sqrt{5}}|})}+\mathcal{C}.

De aici se afla J in functie de t, se revine apoi la relatia (*), se calculeaza I1  in functie de t si, in sfarsit, se gaseste I in functie de x.

Postat în: PRIMITIVE-liceu

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