Question

Realizar las siguientes integrales por partes. (calculo integral). ​

Answer 1

Respuesta:

el numero 3

Explicación paso a paso:

$\int{xe^{x} senx} \, dx$

$u= x$                                 $dv= sen x dx$

$du=dx$                             $v= cos x$

$x(\frac{-1}{2}e^{x} cos(x)+\frac{1}{2}e^{x}sen(x))-\frac{1}{2}\int{-e^{x}cos(x)+ e^{x}sen(x)} \, dx$

$x(\frac{-1}{2}e^{x} cos(x)+\frac{1}{2}e^{x}sen(x))-(-\frac{1}{2}e^{x}cos(x)) \, dx$

$x(\frac{-1}{2}e^{x} cos(x)+\frac{1}{2}e^{x}sen(x))+\frac{1}{2}e^{x}cos(x)\, dx\\+c$

Answer 2

Respuestas y pasos:

1. $\int \:e^{2x}\:\cdot \tan ^{-1}\left(e^x\right)\:dx$

$\mathrm{Aplicar\:integracion\:por\:sustitucion:}\:u=e^x$

$=\int \:u\arctan \left(u\right)du$

$\mathrm{Aplicar\:integracion\:por\:partes:}\:u=\arctan \left(u\right),\:v'=u$

$=\frac{1}{2}u^2\arctan \left(u\right)-\int \frac{u^2}{2\left(u^2+1\right)}du$

$\int \frac{u^2}{2\left(u^2+1\right)}du=\frac{1}{2}\left(-\arctan \left(u\right)+u\right)$

$=\frac{1}{2}u^2\arctan \left(u\right)-\frac{1}{2}\left(-\arctan \left(u\right)+u\right)$

$\mathrm{Sustituir\:en\:la\:ecuacion}\:u=e^x$

$=\frac{1}{2}\left(e^x\right)^2\arctan \left(e^x\right)-\frac{1}{2}\left(-\arctan \left(e^x\right)+e^x\right)$

$\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \left(a^b\right)^c=a^{bc}$

$=\frac{1}{2}e^{2x}\arctan \left(e^x\right)-\frac{1}{2}\left(-\arctan \left(e^x\right)+e^x\right)$

$\mathrm{Agregar\:una\:constante\:a\:la\:solucion}$

$=\frac{1}{2}e^{2x}\arctan \left(e^x\right)-\frac{1}{2}\left(-\arctan \left(e^x\right)+e^x\right)+C$

2. $\int \frac{x^2\cdot \:e^x}{\left(x+2\right)^2}dx$

$\mathrm{Aplicar\:integracion\:por\:partes:}\:u=x^2e^x,\:v'=\frac{1}{\left(x+2\right)^2}$

$=-\frac{e^xx^2}{x+2}-\int \:-e^xxdx$

$=-\frac{e^xx^2}{x+2}-\left(-e^xx+e^x\right)$

$Simplificar$

$=-\frac{e^xx^2}{x+2}+e^xx-e^x$

$\mathrm{Agregar\:una\:constante\:a\:la\:solucion}$

$=-\frac{e^xx^2}{x+2}+e^xx-e^x+C$

3. $\int \:xe^x\sin \left(x\right)dx$

$\mathrm{Aplicar\:integracion\:por\:partes:}\:u=x,\:v'=\sin \left(x\right)e^x$

$=x\left(-\frac{1}{2}e^x\cos \left(x\right)+\frac{1}{2}e^x\sin \left(x\right)\right)-\int \frac{1}{2}\left(-e^x\cos \left(x\right)+e^x\sin \left(x\right)\right)dx$

$\int \frac{1}{2}\left(-e^x\cos \left(x\right)+e^x\sin \left(x\right)\right)dx=-\frac{1}{2}e^x\cos \left(x\right)$

$=x\left(-\frac{1}{2}e^x\cos \left(x\right)+\frac{1}{2}e^x\sin \left(x\right)\right)-\left(-\frac{1}{2}e^x\cos \left(x\right)\right)$

$\mathrm{Simplificar}$

$=x\left(-\frac{1}{2}e^x\cos \left(x\right)+\frac{1}{2}e^x\sin \left(x\right)\right)+\frac{1}{2}e^x\cos \left(x\right)$

$\mathrm{Agregar\:una\:constante\:a\:la\:solucion}$

$=x\left(-\frac{1}{2}e^x\cos \left(x\right)+\frac{1}{2}e^x\sin \left(x\right)\right)+\frac{1}{2}e^x\cos \left(x\right)+C$

4.  $\int \:xe^{-x}\cos \left(2x\right)dx$

$\mathrm{Aplicar\:integracion\:por\:sustitucion:}\:u=-x$

$=\int \:e^uu\cos \left(2u\right)du$

$\mathrm{Aplicar\:integracion\:por\:partes:}\:u=u,\:v'=\cos \left(2u\right)e^u$

$=u\left(\frac{2}{5}e^u\sin \left(2u\right)+\frac{1}{5}e^u\cos \left(2u\right)\right)-\int \frac{1}{5}\left(2e^u\sin \left(2u\right)+e^u\cos \left(2u\right)\right)du$

$\int \frac{1}{5}\left(2e^u\sin \left(2u\right)+e^u\cos \left(2u\right)\right)du=\frac{1}{25}\left(-3e^u\cos \left(2u\right)+4e^u\sin \left(2u\right)\right)$

$=u\left(\frac{2}{5}e^u\sin \left(2u\right)+\frac{1}{5}e^u\cos \left(2u\right)\right)-\frac{1}{25}\left(-3e^u\cos \left(2u\right)+4e^u\sin \left(2u\right)\right)$

$\mathrm{Sustituir\:en\:la\:ecuacion}\:u=-x$ $=\left(-x\right)\left(\frac{2}{5}e^{-x}\sin \left(2\left(-x\right)\right)+\frac{1}{5}e^{-x}\cos \left(2\left(-x\right)\right)\right)-\frac{1}{25}\left(-3e^{-x}\cos \left(2\left(-x\right)\right)+4e^{-x}\sin \left(2\left(-x\right)\right)\right)$$Simplificar$ $=-x\left(\frac{1}{5}e^{-x}\cos \left(2x\right)-\frac{2}{5}e^{-x}\sin \left(2x\right)\right)-\frac{1}{25}\left(-3e^{-x}\cos \left(2x\right)-4e^{-x}\sin \left(2x\right)\right)$

$\mathrm{Agregar\:una\:constante\:a\:la\:solucion}$

$=-x\left(\frac{1}{5}e^{-x}\cos \left(2x\right)-\frac{2}{5}e^{-x}\sin \left(2x\right)\right)-\frac{1}{25}\left(-3e^{-x}\cos \left(2x\right)-4e^{-x}\sin \left(2x\right)\right)+C$

5. $\int \:e^{\:sen^{-1}\left(x\right)}dx$

$\mathrm{Aplicar\:integracion\:por\:partes:}\:u=e^{\arcsin \left(x\right)},\:v'=1$

$=e^{\arcsin \left(x\right)}x-\int \frac{e^{\arcsin \left(x\right)}x}{\sqrt{1-x^2}}dx$

$\int \frac{e^{\arcsin \left(x\right)}x}{\sqrt{1-x^2}}dx=-\frac{1}{2}e^{\arcsin \left(x\right)}\sqrt{1-x^2}+\frac{1}{2}e^{\arcsin \left(x\right)}x$

$=e^{\arcsin \left(x\right)}x-\left(-\frac{1}{2}e^{\arcsin \left(x\right)}\sqrt{1-x^2}+\frac{1}{2}e^{\arcsin \left(x\right)}x\right)$

$Simplificar$

$=\frac{1}{2}e^{\arcsin \left(x\right)}x+\frac{1}{2}e^{\arcsin \left(x\right)}\sqrt{-x^2+1}$

$\mathrm{Agregar\:una\:constante\:a\:la\:solucion}$

$=\frac{1}{2}e^{\arcsin \left(x\right)}x+\frac{1}{2}e^{\arcsin \left(x\right)}\sqrt{-x^2+1}+C$