Why is my evaluation of this integral wrong?

s3a

Nov 2008
624
5
My work is attached and any input would be greatly appreciated!
Thanks in advance!
 

Attachments

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
algebra, algebra, algebra ...

\(\displaystyle u^2+25 \ne (u-5)(u+5)\)


\(\displaystyle \int \frac{e^{4x}}{e^{8x} + 25} \, dx
\)

\(\displaystyle u =e^{4x}\)

\(\displaystyle du = 4e^{4x} \, dx\)

\(\displaystyle \frac{1}{4} \int \frac{4e^{4x}}{e^{8x} + 25} \, dx\)

\(\displaystyle \frac{1}{4} \int \frac{du}{u^2 + 25} \, dx\)

\(\displaystyle \frac{1}{20} \arctan\left(\frac{u}{5}\right) + C
\)

\(\displaystyle \frac{1}{20} \arctan\left(\frac{e^{4x}}{5}\right) + C\)
 
Apr 2010
384
153
Canada
algebra, algebra, algebra ...

\(\displaystyle u^2+25 \ne (u-5)(u+5)\)


\(\displaystyle \int \frac{e^{4x}}{e^{8x} + 25} \, dx
\)

\(\displaystyle u =e^{4x}\)

\(\displaystyle du = 4e^{4x} \, dx\)

\(\displaystyle \frac{1}{4} \int \frac{4e^{4x}}{e^{8x} + 25} \, dx\)

\(\displaystyle \frac{1}{4} \int \frac{du}{u^2 + 25} \, dx\)

\(\displaystyle \frac{1}{20} \arctan\left(\frac{u}{5}\right) + C
\)

\(\displaystyle \frac{1}{20} \arctan\left(\frac{e^{4x}}{5}\right) + C\)
How on god's green earth could you read that sheet!? :0
 

s3a

Nov 2008
624
5
Oh ya! I always make that mistake when I do things fast. Thanks! (and lol sorry for my handwriting)
 
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