Question

Salut, imi puteti spune unde gresesc la aceasta problema...?

Answer 1

$\displaystyle I = \int_{1}^3\dfrac{\ln x}{x^2+3}\, dx \\ \\ I = \int_{1}^{\sqrt 3}\dfrac{\ln x}{x^2+3}\, dx+\int_{\sqrt 3}^{3}\dfrac{\ln x}{x^2+3}\, dx = J + K\\ \\\\ K = \int_{\sqrt 3}^3\dfrac{\ln x}{x^2+3}\, dx\\ \\ u = \dfrac{3}{x} \Rightarrow x = \dfrac{3}{u} \Rightarrow dx = -\dfrac{3}{u^2}\, du\\ x = \sqrt 3\Rightarrow u = \sqrt 3,~~x = 3 \Rightarrow u = 1 \\ \\ K = \int_{\sqrt 3}^1\dfrac{\ln\Big(\dfrac{3}{u}\Big)}{\dfrac{9}{u^2}+3}\cdot \Big(- \dfrac{3}{u^2}\Big)\,du =$

$\displaystyle =\int_{1}^{\sqrt 3}\dfrac{\ln(3)-\ln(u)}{3+u^2}\, du = \ln(3)\int_{1}^{\sqrt 3}\dfrac{1}{u^2+3}\, du - J\\ \\ \\ \Rightarrow I = J + \ln(3)\int_{1}^{\sqrt 3}\dfrac{1}{u^2+3}\, du - J \\ \Rightarrow I = \ln(3)\int_{1}^{\sqrt 3}\dfrac{1}{u^2+3}\, du\\ \\ \Rightarrow I = \ln(3)\cdot \dfrac{1}{\sqrt 3}\arctan\Big(\dfrac{u}{\sqrt 3}\Big)\Bigg|_{1}^{\sqrt 3} \\ \\ \Rightarrow I = \dfrac{ln(3)}{\sqrt 3}\cdot \Bigg(\arctan\Big(\dfrac{\sqrt 3}{\sqrt 3}\Big) - \arctan\Big(\dfrac{1}{\sqrt 3}\Big) \Bigg)$

$\Rightarrow I = \dfrac{ln(3)}{\sqrt 3}\cdot \Big(\dfrac{\pi}{4}-\dfrac{\pi}{6} \Big) \Rightarrow \boxed{I = \dfrac{\pi \ln(3)}{12\sqrt 3}}$