Why is my evaluation of this integral wrong?

s3a

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

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skeeter

MHF Helper
algebra, algebra, algebra ...

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

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

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

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

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

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

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

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

AllanCuz

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

skeeter

MHF Helper
How on god's green earth could you read that sheet!? :0
I decipher hieroglyphs in my spare time ... (Smirk)

s3a

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