# nice integral

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• June 5th 2009, 12:24 PM
a.h.m.a.d
nice integral
find :
• June 5th 2009, 12:48 PM
Amer
Quote:

Originally Posted by a.h.m.a.d
find :

$\int_{1}^{e}(x+1)e^x lnx dx$

$\int_{1}^{e}(xe^x ln(x)+e^x (lnx )dx$

the second integral let
$du=dx,,u=x....v=e^x lnx...dv=e^x lnx + \frac{e^x}{x}$

$\int_{1}^{e}(xe^x ln(x))dx+\int_{1}^{e}e^x (lnx )dx$

$\int_{1}^{e}(xe^x ln(x))dx+xe^x ln(x)-\int_{1}^{e} x(e^x lnx +\frac{e^x}{x})dx$

$\int_{1}^{e}(xe^x ln(x))dx+xe^x ln(x)-\int_{1}^{e} xe^x ln(x) dx+\int_{1}^{e} e^x dx$

$xe^x ln(x)+\int_{1}^{e} e^x dx$

$xe^x ln(x) +e^x$

now just substitute
• June 5th 2009, 12:56 PM
a.h.m.a.d
Amer (Clapping)
• June 5th 2009, 01:26 PM
Soroban
Hello, a.h.m.a.d!

Quote:

Find: . $\int^e_1(x+1)\,e^x\ln x\,dx$
Integration by parts . . .

. . $\begin{array}{ccccccc}u &=& \ln x & & dv &=& (x+1)e^x\,dx \\ du &=& \frac{dx}{x} & & v &=& xe^x\;\;{\color{red}*} \end{array}$

We have: . $xe^x\ln x - \int e^x\,dx \;=\;xe^x\ln x - e^x\,\bigg]^e_1 \quad\hdots\;\text{ etc.}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

* . To integrate: . $\int(x+1)e^x\,dx$

. . . By parts: . $\begin{array}{ccccccc} u &=& x+1 & & dv &=& e^x\,dx \\ du &=& dx & & v &=& e^x \end{array}$

. . . And we have: . $(x+1)e^x - \int e^x\,dx \;\;=\;\;(x+1)e^x - e^x$

. . . . . . $=\;\;xe^x + e^x - e^x \;\;=\;\;xe^x$