Question

Alguien me puede ayudar a resolver esta integral? Lo he intentado muchas veces y no me da :(​

Answer 1

SOLUCIÓN

♛ HØlα!! ✌

$\int\limits{\dfrac{dx}{\sqrt{x}(4-3\sqrt{x})}} \,\\\\\\\mathrm{Cambio \: de\: variable\: \sqrt{x} = u \Rightarrow \boxed{dx =2udu}}\\\\\\\int\limits{\dfrac{dx}{u(4-3u)}} \\\\\\\int\limits{\dfrac{2udu}{u(4-3u)}}\\\\\\\int\limits{\dfrac{2du}{(4-3u)}}\\\\\\2\int\limits{\dfrac{du}{(4-3u)}}\\\\\\\mathrm{Sustituci\'on \:v =4-3u\Rightarrow dv = -3du\Rightarrow \boxed{du=\dfrac{dv}{-3}}}\\\\\\2\int\limits{\dfrac{\frac{dv}{-3}}{v}}\\\\\\2\int\limits{\dfrac{dv}{-3v}}$

$-\dfrac{2}{3}\int\limits{\dfrac{dv}{v}}\\\\\\-\dfrac{2}{3}\ln|v|+C\\\\\\\mathrm{Sustituimos \:v}\\\\\\-\dfrac{2}{3}\ln|4-3u|+C\\\\\\\mathrm{Sustituimos \:u}\\\\\\\boxed{\boxed{\boldsymbol{-\dfrac{2}{3}\ln|4-3(\sqrt{x})|+C}}}$