# integration again

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• April 30th 2010, 04:34 AM
Adre5
integration again
hmm can someone please check if this is correct for the next line:

$\int (e^x - 1)^6 e^x dx$
=
$\frac{x(x-1)^7 e^x }{7} + C$

??
• April 30th 2010, 04:45 AM
sa-ri-ga-ma
Quote:

Originally Posted by Adre5
hmm can someone please check if this is correct for the next line:

$\int (e^x - 1)^6 e^x dx$
=
$\frac{x(x-1)^7 e^x }{7} + C$

??

No. If you substitute e^x as t, the e^x*dx = dt.
The integration becomes
Int(t-1)^6*dt. And it is equal to
(t-1)^7/7.
Now substitute for t to get the final result.