Question

Cum rezolv asta? Ex 2 punctual b

Answer 1

$\displaystyle I_n = \int \dfrac{x^n}{\sqrt{x^2+1}}\, dx= \int x^{n-1}\cdot \dfrac{x}{\sqrt{x^2+1}}\, dx = \\ \\ = \int x^{n-1}\cdot (\sqrt{x^2+1})'\, dx= \\ \\ = x^{n-1}\cdot \sqrt{x^2+1}-(n-1)\int x^{n-2}\cdot \sqrt{x^2+1}\, dx = \\ \\=  x^{n-1}\cdot \sqrt{x^2+1}-(n-1)\int x^{n-2}\cdot \dfrac{x^2+1}{\sqrt{x^2+1}}\, dx=\\ \\ = x^{n-1}\cdot \sqrt{x^2+1}-(n-1)\int \dfrac{x^n}{\sqrt{x^2+1}}+\dfrac{x^{n-2}}{\sqrt{x^2+1}}\, dx=$

$\displaystyle = x^{n-1}\cdot \sqrt{x^2+1}-(n-1)(I_n+I_{n-2}) \\ \\ I_n = x^{n-1}\cdot \sqrt{x^2+1}-(n-1)(I_n+I_{n-2})\\ \\ I_n = x^{n-1}\cdot \sqrt{x^2+1}-(n-1)I_n -(n-1)I_{n-2}\\ \\I_n +(n-1)I_{n} =x^{n-1}\cdot \sqrt{x^2+1}-(n-1)I_{n-2} \\ \\ nI_{n} =x^{n-1}\cdot \sqrt{x^2+1} -(n-1)I_{n-2} \\ \\ I_{n} = \dfrac{x^{n-1}\cdot \sqrt{x^2+1} -(n-1)I_{n-2}}{n}$