Question

AM170
Ce aș putea să fac la integrala asta?

Answer 1

Salut,

Problema se rezolvă cu metoda integrării prin părți, uite așa:

$\int_{\sqrt3/3}^{\sqrt3}arctg\dfrac{1}xdx=\int_{\sqrt3/3}^{\sqrt3}x^{'}\cdot arctg\dfrac{1}xdx=\\\\=x\cdot arctg\dfrac{1}x\Bigg{|}_{\sqrt3/3}^{\sqrt3}-\int_{\sqrt3/3}^{\sqrt3}x\cdot\dfrac{1}{1+\dfrac{1}{x^2}}\cdot\left(-\dfrac{1}{x^2}\right)dx=\sqrt3\cdot arctg\dfrac{1}{\sqrt3}-\dfrac{\sqrt3}{3}\cdot arctg\dfrac{1}{\dfrac{\sqrt3}3}+\\\\+\int_{\sqrt3/3}^{\sqrt3}\dfrac{1}{x+\dfrac{1}x}dx=\sqrt3\cdot arctg\dfrac{\sqrt3}{3}-\dfrac{\sqrt3}{3}\cdot arctg\sqrt3+\int_{\sqrt3/3}^{\sqrt3}\dfrac{x}{x^2+1}dx=\\\\=\sqrt3\cdot\dfrac{\pi}6-\dfrac{\sqrt3}{3}\cdot\dfrac{\pi}3+\dfrac{1}{2}\cdot\int_{\sqrt3/3}^{\sqrt3}\dfrac{2x}{x^2+1}dx=\dfrac{\sqrt3\cdot\pi}{18}+\dfrac{1}{2}\cdot ln|x^2+1|\Bigg{|}_{\sqrt3/3}^{\sqrt3}=\\\\=\dfrac{\sqrt3\cdot\pi}{18}+\dfrac{ln|4|-ln\left|\dfrac{1}{3}+1\right|}{2}=\dfrac{\sqrt3\cdot\pi}{18}+\dfrac{ln4-ln\left(\dfrac{4}{3}\right)}{2}=\\\\=\dfrac{\sqrt3\cdot\pi}{18}+\dfrac{ln4-ln4+ln3}{2}=\dfrac{\sqrt3\cdot\pi}{18}+\dfrac{ln3}{2}=\dfrac{\pi}{6\sqrt3}+\dfrac{ln3}{2}.$

Green eyes.