# its a solved integration

• October 13th 2009, 10:28 AM
alex83
its a solved integration
i need an explanation to the underlined answers pls (check attached picture)

i got different answer for the first one

i got : e^(lny+y)-e^y=y

thnx
• November 7th 2009, 10:45 AM
Moo
Hello,

Strange that no one replied to this...

$[e^{x+y}]_0^{\ln(y)}=e^{\ln(y)+y}-e^{0+y}=e^{\ln(y)}e^y-e^y=ye^y-e^y=(y-1)e^y$

then they just performed an integration by parts :)