Question

sa se calculeze integrala din 2x ln radicaldin 4 de ordinul 3 -ln 3 ori 9 la x

Answer 1

$\int{2x\ln 4^{\frac{1}{3}}}-\ln 3 \cdot 9^x \ \ dx=\\ \\ \\ =2\ln 4^{\frac{1}{3}}\int{x\ dx}-\ln 3 \int{9^x\ dx}= \\ \\ \\ =2\ln 2^{\frac{2}{3}}\cdot\dfrac{x^2}{2}-\ln 3 \cdot \dfrac{9^x}{\ln 9}+C=\\ \\ \\ =\dfrac{2}{3}\ln 2\cdot x^2-\ln 3\cdot \dfrac{9^x}{2\ln 3}+C=\\ \\ \\ =\dfrac{2\ln 2}{3}x^2-\dfrac{9^x}{2}+C.$