preciso de ajuda alguem pode me ajudar com este exercicio
3) Sabendo que para se resolver uma integral por parte se deve utilizar a seguinte formula:∫u . dv = u . v -∫ v . du
Diante disso calcule a seguinte integral utilizando o método da integração por partes.
a) ∫ xlnxdx =
Soluções para a tarefa
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Boa tarde Silva!
Solução!
![\displaystyle \int (xlnx)dx\\\\\\
Fazendo!\\\\\\
u=lnx~~~~dv=xdx\\\\\
du= \frac{1}{x}dx~~~~~~v= \frac{ x^{2} }{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cint+%28xlnx%29dx%5C%5C%5C%5C%5C%5C%0AFazendo%21%5C%5C%5C%5C%5C%5C%0Au%3Dlnx%7E%7E%7E%7Edv%3Dxdx%5C%5C%5C%5C%5C%0Adu%3D+%5Cfrac%7B1%7D%7Bx%7Ddx%7E%7E%7E%7E%7E%7Ev%3D+%5Cfrac%7B+x%5E%7B2%7D+%7D%7B2%7D++ "\displaystyle \int (xlnx)dx\\\\\\
Fazendo!\\\\\\
u=lnx~~~~dv=xdx\\\\\
du= \frac{1}{x}dx~~~~~~v= \frac{ x^{2} }{2}")
![\boxed{\displaystyle \int udv=uv-\displaystyle\int vdu}\\\\\\\\
\displaystyle \int (xlnx)=lnx. \frac{ x^{2} }{2}- \displaystyle \int \frac{ x^{2} }{2}.\frac{1}{x}dx \\\\\\\
Simplificando!\\\\\\\
\displaystyle \int (xlnx)=lnx. \frac{ x^{2} }{2}- \displaystyle \int \frac{ x }{2}dx \\\\\\\
\displaystyle \int (xlnx)=lnx. \frac{ x^{2} }{2}- \frac{1}{2} \displaystyle \int (x)dx \\\\\\\](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cdisplaystyle+%5Cint+udv%3Duv-%5Cdisplaystyle%5Cint+vdu%7D%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Cdisplaystyle+%5Cint+%28xlnx%29%3Dlnx.+%5Cfrac%7B+x%5E%7B2%7D+%7D%7B2%7D-+%5Cdisplaystyle+%5Cint++%5Cfrac%7B+x%5E%7B2%7D+%7D%7B2%7D.%5Cfrac%7B1%7D%7Bx%7Ddx+%5C%5C%5C%5C%5C%5C%5C%0ASimplificando%21%5C%5C%5C%5C%5C%5C%5C%0A%5Cdisplaystyle+%5Cint+%28xlnx%29%3Dlnx.+%5Cfrac%7B+x%5E%7B2%7D+%7D%7B2%7D-+%5Cdisplaystyle+%5Cint++%5Cfrac%7B+x+%7D%7B2%7Ddx+%5C%5C%5C%5C%5C%5C%5C%0A%5Cdisplaystyle+%5Cint+%28xlnx%29%3Dlnx.+%5Cfrac%7B+x%5E%7B2%7D+%7D%7B2%7D-++%5Cfrac%7B1%7D%7B2%7D+%5Cdisplaystyle+%5Cint++%28x%29dx+%5C%5C%5C%5C%5C%5C%5C%0A%0A%0A "\boxed{\displaystyle \int udv=uv-\displaystyle\int vdu}\\\\\\\\
\displaystyle \int (xlnx)=lnx. \frac{ x^{2} }{2}- \displaystyle \int \frac{ x^{2} }{2}.\frac{1}{x}dx \\\\\\\
Simplificando!\\\\\\\
\displaystyle \int (xlnx)=lnx. \frac{ x^{2} }{2}- \displaystyle \int \frac{ x }{2}dx \\\\\\\
\displaystyle \int (xlnx)=lnx. \frac{ x^{2} }{2}- \frac{1}{2} \displaystyle \int (x)dx \\\\\\\")
![\displaystyle \int (xlnx)=lnx. \frac{ x^{2} }{2}- \frac{1}{2}. \frac{ x^{2} }{2} +c \\\\\\\
\displaystyle \int (xlnx)=lnx. \frac{ x^{2} }{2}- \frac{ x^{2} }{4} +c \\\\\\\\\\\\
\boxed{Resposta:\displaystyle \int (xlnx)dx=lnx. \frac{ x^{2} }{2}- \frac{ x^{2} }{4} +c}](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cint+%28xlnx%29%3Dlnx.+%5Cfrac%7B+x%5E%7B2%7D+%7D%7B2%7D-++%5Cfrac%7B1%7D%7B2%7D.+%5Cfrac%7B+x%5E%7B2%7D+%7D%7B2%7D++%2Bc+%5C%5C%5C%5C%5C%5C%5C%0A%5Cdisplaystyle+%5Cint+%28xlnx%29%3Dlnx.+%5Cfrac%7B+x%5E%7B2%7D+%7D%7B2%7D-+++%5Cfrac%7B+x%5E%7B2%7D+%7D%7B4%7D++%2Bc+%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BResposta%3A%5Cdisplaystyle+%5Cint+%28xlnx%29dx%3Dlnx.+%5Cfrac%7B+x%5E%7B2%7D+%7D%7B2%7D-+++%5Cfrac%7B+x%5E%7B2%7D+%7D%7B4%7D++%2Bc%7D%0A "\displaystyle \int (xlnx)=lnx. \frac{ x^{2} }{2}- \frac{1}{2}. \frac{ x^{2} }{2} +c \\\\\\\
\displaystyle \int (xlnx)=lnx. \frac{ x^{2} }{2}- \frac{ x^{2} }{4} +c \\\\\\\\\\\\
\boxed{Resposta:\displaystyle \int (xlnx)dx=lnx. \frac{ x^{2} }{2}- \frac{ x^{2} }{4} +c}")
Boa tarde!
Bons estudos!
Solução!
Boa tarde!
Bons estudos!
silvaseixas1:
obrigado pela a ajuda.!
Respondido por
0
Vejamos:
u = x du = x²/2
v = lnxdx dv = 1/xdx
∫(xlnx)dx = x²lnx-∫(xlnx-x)dx=x²lnx-∫xdx = 2∫xlnx=x²lnx - x²/2= x²/2(lnx)-x²/4
u = x du = x²/2
v = lnxdx dv = 1/xdx
∫(xlnx)dx = x²lnx-∫(xlnx-x)dx=x²lnx-∫xdx = 2∫xlnx=x²lnx - x²/2= x²/2(lnx)-x²/4
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