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Resolvamos por cambio de variable:
Si hacemos: u=lnx , entonces: x=e^u --> dx = e^u*du
Reemplazando:
![\int \frac{lnx}{ \sqrt{x} } dx= \int \frac{u}{e^{u/2}} e^udu = \int ue^{u/2}du \int \frac{lnx}{ \sqrt{x} } dx= \int \frac{u}{e^{u/2}} e^udu = \int ue^{u/2}du](https://tex.z-dn.net/?f=+%5Cint+%5Cfrac%7Blnx%7D%7B+%5Csqrt%7Bx%7D+%7D+dx%3D+%5Cint+%5Cfrac%7Bu%7D%7Be%5E%7Bu%2F2%7D%7D+e%5Eudu+%3D+%5Cint+ue%5E%7Bu%2F2%7Ddu)
Integración por partes: U=u --> dU=du ; d V=e^(u/2)du --> V = 2*e^(u/2)
![\int \frac{lnx}{ \sqrt{x} } dx = \int \frac{u}{e^{u/2}} e^udu = U*V-\int VdU
\ \
\int \frac{lnx}{ \sqrt{x} } dx = \int \frac{u}{e^{u/2}} e^udu = u*2e^{u/2}-\int {2e^{u/2}du}
\ \
\int \frac{lnx}{ \sqrt{x} } dx = \int \frac{u}{e^{u/2}} e^udu = 2ue^{u/2}-4e^{u/2} + C
\int \frac{lnx}{ \sqrt{x} } dx = \int \frac{u}{e^{u/2}} e^udu = U*V-\int VdU
\ \
\int \frac{lnx}{ \sqrt{x} } dx = \int \frac{u}{e^{u/2}} e^udu = u*2e^{u/2}-\int {2e^{u/2}du}
\ \
\int \frac{lnx}{ \sqrt{x} } dx = \int \frac{u}{e^{u/2}} e^udu = 2ue^{u/2}-4e^{u/2} + C](https://tex.z-dn.net/?f=%5Cint+%5Cfrac%7Blnx%7D%7B+%5Csqrt%7Bx%7D+%7D+dx+%3D+%5Cint+%5Cfrac%7Bu%7D%7Be%5E%7Bu%2F2%7D%7D+e%5Eudu+%3D+U%2AV-%5Cint+VdU%0A%0A%5C+%5C%0A%0A%5Cint+%5Cfrac%7Blnx%7D%7B+%5Csqrt%7Bx%7D+%7D+dx+%3D+%5Cint+%5Cfrac%7Bu%7D%7Be%5E%7Bu%2F2%7D%7D+e%5Eudu+%3D+u%2A2e%5E%7Bu%2F2%7D-%5Cint+%7B2e%5E%7Bu%2F2%7Ddu%7D%0A%0A%5C+%5C%0A%0A%5Cint+%5Cfrac%7Blnx%7D%7B+%5Csqrt%7Bx%7D+%7D+dx+%3D+%5Cint+%5Cfrac%7Bu%7D%7Be%5E%7Bu%2F2%7D%7D+e%5Eudu+%3D+2ue%5E%7Bu%2F2%7D-4e%5E%7Bu%2F2%7D+%2B+C%0A)
pero: u=lnx, reemplazando, obtenemos:
![\int \frac{lnx}{ \sqrt{x} } dx = 2(lnx)e^{(lnx)/2}-4e^{(lnx)/2} + C
\ \
\int \frac{lnx}{ \sqrt{x} } dx = 2(lnx)\sqrt{e^{(lnx)}}-4\sqrt{e^{(lnx)}} + C
\ \
\int \frac{lnx}{ \sqrt{x} } dx = 2(lnx)\sqrt{x}-4\sqrt{x}} + C
\ \
\boxed{ \int \frac{lnx}{ \sqrt{x} } dx = 2\sqrt{x}(lnx-2) + C } \int \frac{lnx}{ \sqrt{x} } dx = 2(lnx)e^{(lnx)/2}-4e^{(lnx)/2} + C
\ \
\int \frac{lnx}{ \sqrt{x} } dx = 2(lnx)\sqrt{e^{(lnx)}}-4\sqrt{e^{(lnx)}} + C
\ \
\int \frac{lnx}{ \sqrt{x} } dx = 2(lnx)\sqrt{x}-4\sqrt{x}} + C
\ \
\boxed{ \int \frac{lnx}{ \sqrt{x} } dx = 2\sqrt{x}(lnx-2) + C }](https://tex.z-dn.net/?f=%5Cint+%5Cfrac%7Blnx%7D%7B+%5Csqrt%7Bx%7D+%7D+dx+%3D+2%28lnx%29e%5E%7B%28lnx%29%2F2%7D-4e%5E%7B%28lnx%29%2F2%7D+%2B+C+%0A%0A%5C+%5C+%0A%0A%5Cint+%5Cfrac%7Blnx%7D%7B+%5Csqrt%7Bx%7D+%7D+dx+%3D+2%28lnx%29%5Csqrt%7Be%5E%7B%28lnx%29%7D%7D-4%5Csqrt%7Be%5E%7B%28lnx%29%7D%7D+%2B+C+%0A%0A%5C+%5C%0A%0A%5Cint+%5Cfrac%7Blnx%7D%7B+%5Csqrt%7Bx%7D+%7D+dx+%3D+2%28lnx%29%5Csqrt%7Bx%7D-4%5Csqrt%7Bx%7D%7D+%2B+C%0A%0A+%5C+%5C%0A%0A+%5Cboxed%7B+%5Cint+%5Cfrac%7Blnx%7D%7B+%5Csqrt%7Bx%7D+%7D+dx+%3D+2%5Csqrt%7Bx%7D%28lnx-2%29+%2B+C+%7D+)
Saludos!!
Si hacemos: u=lnx , entonces: x=e^u --> dx = e^u*du
Reemplazando:
Integración por partes: U=u --> dU=du ; d V=e^(u/2)du --> V = 2*e^(u/2)
pero: u=lnx, reemplazando, obtenemos:
Saludos!!
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