Question

Calculati:
Am mevoie de o indicatie.
Multumesc!!!

Answer 1

Am atașat rezolvarea!

Answer 2

$\text{Formula lui King:}\\ \\ \displaystyle \int_{a}^bf(x)\, dx = \int_{a}^bf(a+b-x)\, dx$

$\displaystyle I =\int_{-1}^1\ln\Big(x+\sqrt{1+x^2}\Big)\, dx \\ \\I = \int_{-1}^1 \ln\Big((-1+1-x)+\sqrt{1+(-1+1-x)^2}\Big)\, dx\\ \\ I = \int_{-1}^1\ln\Big(-x+\sqrt{1+x^2}\Big)\,dx\\ \\ I+I = \int_{-1}^1\Big[\ln\Big(x+\sqrt{1+x^2}\Big)+\ln\Big(-x+\sqrt{1+x^2}\Big)\Big]\, dx \\ \\ 2I = \int_{-1}^1\ln \Big[(\sqrt{1+x^2}+x)(\sqrt{1+x^2}-x)\Big]\, dx \\ \\ 2I = \int_{-1}^1\ln (1+x^2-x^2)\, dx \\ \\ 2I = \int_{-1}^1 \ln 1\, dx \\ \\ 2I = 0 \\ \\ \Rightarrow \boxed{I = 0}$

Se mai poate face si asa:

$\text{Daca f(x) este functie impara.}\\ \\ \displaystyle \Rightarrow \int_{-a}^{a}f(x)\, dx = 0 \\ \\ I = \int_{-1}^1\ln(x+\sqrt{x^2+1})\, dx \\ \\ f(x) = \ln(x+\sqrt{x^2+1})\\ f(-x) = \ln(-x+\sqrt{x^2+1}) = -\ln\Big((-x+\sqrt{x^2+1})^{-1}\Big) = \\ \\ =-\ln \dfrac{1}{-x+\sqrt{x^2+1}} = -\ln \dfrac{\sqrt{x^2+1}+x}{x^2+1-x^2} = -\ln (x+\sqrt{x^2+1)} =\\ \\ = -f(x) \Rightarrow f(x)\rightarrow \text{impar} \\ \\ \Rightarrow \boxed{I = 0}$