Question

Determine as integrais

Answer 1

Resposta:

OLÁ

VAMOS A SUA PERGUNTA:⇒⇒

A:=====>

$\sf \displaystyle\int[2cos(3x)+x\sqrt{x^2+1}]dx$

$\sf \displaystyle\int2cos(3x)+x\sqrt{x^2+1}~dx$

$\sf =\displaystyle\int2cos(3x)dx+\displaystyle\int x\sqrt{x^2+1}~dx$

$\sf \displaystyle\int2cos(3x)dx=\dfrac{2}{3}sin(3x)$

$\sf \displaystyle\int x\sqrt{x^2+1} ~ dx=\dfrac{1}{3}(x^2+1)^\dfrac{3}{2}$

$\sf =\dfrac{2}{3}sin(3x)+\dfrac{1}{3} (x^2+1)^\dfrac{3}{2}$

$\boxed{\bold{\displaystyle{\clubsuit\ \spadesuit\ \maltese\ \sf \red{ =\dfrac{2}{3}sin(3x)+\dfrac{1}{3} (x^2+1)^\dfrac{3}{2}}}+C}}}\ \checkmark$← RESPOSTA.

B:=====>

$\sf \displaystyle\int\dfrac{x^2+2x}{\sqrt{x^3+3x^2+1} }dx$

$=\sf \displaystyle\int\dfrac{1}{3\sqrt{u} }~du$

$\sf=\dfrac{1}{3}\times \displaystyle\int\dfrac{1}{\sqrt{u} }~du$

$\sf=\dfrac{1}{3}\times \displaystyle\int\dfrac{1}{u^\dfrac{1}{2} }~du$

$\sf=\dfrac{1}{3}\times \displaystyle\int u^-^\dfrac{1}{2}}du$

$\sf =\dfrac{1}{3 }\times\dfrac{u-\dfrac{1}{2}+1 }{-\dfrac{1}{2}+1 }$

$\sf =\dfrac{1}{3 }\times\dfrac{(x^3+3x^2+1)-\dfrac{1}{2}+1 }{-\dfrac{1}{2}+1 }$

$\sf =\dfrac{2}{3 }\sqrt{x^3+3x^2+1}$

$\boxed{\bold{\displaystyle{\clubsuit\ \spadesuit\ \maltese\ \sf \green{=\dfrac{2}{3 }\sqrt{x^3+3x^2+1}+C}}}}\ \checkmark$← RESPOSTA.

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