Sa se gaseasca o relatie de recurenta pentru \( I_n=\int\frac{x^n}{\sqrt{x^2+1}}dx,n\in\mathbb{N} \)
si apoi sa se determine \( I_0,I_1 \) si \( I_2 \)
Sirb Vasile, Zalau
Concurs "Teodor Topan" - problema 2
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Kunihiko Chikaya
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\( I_0=\int \frac{1}{\sqrt{x^2+1}}\ dx=\ln |x+\sqrt{x^2+1}|+C. \)
\( I_1=\int \frac{x}{\sqrt{x^2+1}}\ dx=\frac{1}{2}\int \frac{1}{\sqrt{x^2+1}}d(x^2+1)=\sqrt{x^2+1}+C. \)
\( I_2=\int \frac{x^2}{\sqrt{x^2+1}}\ dx=\int \frac{(x^2+1)-1}{\sqrt{x^2+1}}\ dx
=\int \left(\sqrt{x^2+1}-\frac{1}{\sqrt{x^2+1}}\right)\ dx
=\int \sqrt{x^2+1}dx-I_0 \)
\( =x\sqrt{x^2+1}-\int_0^1 x\cdot \frac{2x}{2\sqrt{x^2+1}}dx-I_0 \)
\( =x\sqrt{x^2+1}-I_2-I_0 \)
\( \therefore I_2=\frac{1}{2}(x\sqrt{x^2+1}-\ln |x+\sqrt{x^2+1}|)+C. \)
\( I_1=\int \frac{x}{\sqrt{x^2+1}}\ dx=\frac{1}{2}\int \frac{1}{\sqrt{x^2+1}}d(x^2+1)=\sqrt{x^2+1}+C. \)
\( I_2=\int \frac{x^2}{\sqrt{x^2+1}}\ dx=\int \frac{(x^2+1)-1}{\sqrt{x^2+1}}\ dx
=\int \left(\sqrt{x^2+1}-\frac{1}{\sqrt{x^2+1}}\right)\ dx
=\int \sqrt{x^2+1}dx-I_0 \)
\( =x\sqrt{x^2+1}-\int_0^1 x\cdot \frac{2x}{2\sqrt{x^2+1}}dx-I_0 \)
\( =x\sqrt{x^2+1}-I_2-I_0 \)
\( \therefore I_2=\frac{1}{2}(x\sqrt{x^2+1}-\ln |x+\sqrt{x^2+1}|)+C. \)