Question

Essa integral por partes ∫ t ln(2t) dt , alguém pode resolver para mim passo a passo?

Answer 1

Boa tarde!

Solução!

$\displaystyle \int (t.ln(2t)dt$

Modelo para substituição!

$\boxed{\displaystyle \int u.dv=uv-\displaystyle \int vdu}$

Retomando a integral em questão!

$u=ln(2t)~~~~~~dv=t\\\\\\ du= \dfrac{1}{t}dt~~~~~~v= \dfrac{t^{2} }{2}\\\\\\ \displaystyle \int (t.ln(2t)=ln(2t). \frac{t^{2} }{2}- \displaystyle \int (\frac{t ^{2} }{2}. \frac{1}{t}) dt\\\\\\\ \displaystyle \int (t.ln(2t)=ln(2t). \frac{t^{2} }{2}- \frac{1}{2} \displaystyle \int (\frac{t ^{2} }{t}) dt\\\\\\\ \displaystyle \int (t.ln(2t)=ln(2t). \frac{t^{2} }{2}- \frac{1}{2} \displaystyle \int (t) dt\\\\\\\ \displaystyle \int (t) dt= \frac{t^{2} }{2}+c$


$\displaystyle \int (t.ln(2t)=ln(2t). \frac{t^{2} }{2}- \frac{1}{2} . \frac{t^{2} }{2}+c\\\\\\\\\ \displaystyle \int (t.ln(2t)=ln(2t). \frac{t^{2} }{2}- \frac{t^{2} }{4}+c\\\\\\\\\ \boxed{Resposta:\displaystyle \int (t.ln(2t)=ln(2t). \frac{t^{2} }{2}- \frac{t^{2} }{4}+c}$


Boa tarde!
Bons estudos!