S s3a Nov 2008 624 5 May 19, 2010 #1 My work is attached and any input would be greatly appreciated! Thanks in advance! Attachments mywork.jpg 457.3 KB Views: 16

skeeter MHF Helper Jun 2008 16,216 6,764 North Texas May 19, 2010 #2 algebra, algebra, algebra ... \(\displaystyle u^2+25 \ne (u-5)(u+5)\) \(\displaystyle \int \frac{e^{4x}}{e^{8x} + 25} \, dx \) \(\displaystyle u =e^{4x}\) \(\displaystyle du = 4e^{4x} \, dx\) \(\displaystyle \frac{1}{4} \int \frac{4e^{4x}}{e^{8x} + 25} \, dx\) \(\displaystyle \frac{1}{4} \int \frac{du}{u^2 + 25} \, dx\) \(\displaystyle \frac{1}{20} \arctan\left(\frac{u}{5}\right) + C \) \(\displaystyle \frac{1}{20} \arctan\left(\frac{e^{4x}}{5}\right) + C\)

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 Apr 2010 384 153 Canada May 19, 2010 #3 skeeter said: algebra, algebra, algebra ... \(\displaystyle u^2+25 \ne (u-5)(u+5)\) \(\displaystyle \int \frac{e^{4x}}{e^{8x} + 25} \, dx \) \(\displaystyle u =e^{4x}\) \(\displaystyle du = 4e^{4x} \, dx\) \(\displaystyle \frac{1}{4} \int \frac{4e^{4x}}{e^{8x} + 25} \, dx\) \(\displaystyle \frac{1}{4} \int \frac{du}{u^2 + 25} \, dx\) \(\displaystyle \frac{1}{20} \arctan\left(\frac{u}{5}\right) + C \) \(\displaystyle \frac{1}{20} \arctan\left(\frac{e^{4x}}{5}\right) + C\) Click to expand... How on god's green earth could you read that sheet!? :0

skeeter said: algebra, algebra, algebra ... \(\displaystyle u^2+25 \ne (u-5)(u+5)\) \(\displaystyle \int \frac{e^{4x}}{e^{8x} + 25} \, dx \) \(\displaystyle u =e^{4x}\) \(\displaystyle du = 4e^{4x} \, dx\) \(\displaystyle \frac{1}{4} \int \frac{4e^{4x}}{e^{8x} + 25} \, dx\) \(\displaystyle \frac{1}{4} \int \frac{du}{u^2 + 25} \, dx\) \(\displaystyle \frac{1}{20} \arctan\left(\frac{u}{5}\right) + C \) \(\displaystyle \frac{1}{20} \arctan\left(\frac{e^{4x}}{5}\right) + C\) Click to expand... How on god's green earth could you read that sheet!? :0

skeeter MHF Helper Jun 2008 16,216 6,764 North Texas May 19, 2010 #4 AllanCuz said: How on god's green earth could you read that sheet!? :0 Click to expand... I decipher hieroglyphs in my spare time ... (Smirk) Reactions: HallsofIvy, Archie Meade, s3a and 1 other person

AllanCuz said: How on god's green earth could you read that sheet!? :0 Click to expand... I decipher hieroglyphs in my spare time ... (Smirk)

S s3a Nov 2008 624 5 May 19, 2010 #5 Oh ya! I always make that mistake when I do things fast. Thanks! (and lol sorry for my handwriting) Reactions: Archie Meade