# Thread: Why is my evaluation of this integral wrong?

1. ## Why is my evaluation of this integral wrong?

My work is attached and any input would be greatly appreciated!

2. algebra, algebra, algebra ...

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

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

$\displaystyle u =e^{4x}$

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

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

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

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

$\displaystyle \frac{1}{20} \arctan\left(\frac{e^{4x}}{5}\right) + C$

3. Originally Posted by skeeter
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

4. Originally Posted by AllanCuz
How on god's green earth could you read that sheet!? :0
I decipher hieroglyphs in my spare time ...

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