Question

quem souber ajude. .............

Answer 1

$\int\limits{ \sqrt{x}* lnx} \, dx$

Vamos usar a regrinha do "LIATE"

L = Função logaritma

A = Aritmética.

Então,

u = Lnx  e   Dv = √x*dx

$\\ u = Lnx \\ \\ \frac{du}{dx} = \frac{d(Lnx)}{dx} \\ \\ \frac{du}{dx} = \frac{1}{x} \\ \\ du = \frac{dx}{x}$
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$\\ dv = \sqrt{x} dx \\ \\ dv = x^ \frac{1}{2} dx \\ \\ \int\limits\, dv = \int\limits x^ \frac{1}{2} \, dx \\ \\ v = \frac{x^ \frac{1}{2} ^+^1}{ \frac{1}{2}+1 } \\ \\ v = \frac{x^ \frac{3}{2} }{ \frac{3}{2} } \\ \\ v = \frac{2x^ \frac{3}{2} }{3}$

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Usando integração por partes:


$\\ \int\limits \sqrt{x} Lnx \, dx = uv- \int\limits v\, du \\ \\ = \frac{2x^ \frac{3}{2} }{3} Ln(x)- \int\limits \frac{2x^ \frac{3}{2} }{3}\, \frac{dx}{x} \\ \\ = \frac{2x^ \frac{3}{2} }{3} Ln(x)- \int\limits \frac{2x^ \frac{3}{2} ^-^1}{3}\, dx \\ \\ = \frac{2x^ \frac{3}{2} }{3} Ln(x)- \int\limits \frac{2x^ \frac{1}{2} }{3}\, dx \\ \\ = \frac{2x^ \frac{3}{2} }{3} Ln(x)- \frac{2}{3} * \frac{x^ \frac{1}{2}^+^1 }{\frac{1}{2}^+^1} +C$

$\\ = \frac{2x^ \frac{3}{2} }{3} Ln(x)- \frac{2}{3} * \frac{x^ \frac{3}{2} }{\frac{3}{2}} +C \\ \\ = \frac{2x^ \frac{3}{2} }{3} Ln(x)- \frac{2}{3} * \frac{2x^ \frac{3}{2} }{3} +C$

$\\ = \frac{2x^ \frac{3}{2} }{3} Ln(x)- \frac{4}{9} *x^ \frac{3}{2} +C$