Hi, Could someone please tell me if this is correct? int. e^2x/sq.2+e^x int. du/a^2+b^2=1/a arctanu/a+c u=2+e^x du=e^xdx int. e^2x/sq. 2+e^x=1/sq.2 arctan 2+e^x/sq. 2+c Thank you
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It's about $\displaystyle \displaystyle\int\frac{e^{2x}}{2^2+e^x}dx$ or $\displaystyle \displaystyle\int\frac{e^{2x}}{\sqrt{2+e^x}}dx$ ?
I think it's the second one. Just set $\displaystyle u=\sqrt{2+e^x}$
This one Sorry for the confusion
Just set Do I just plug this into the forumula or do I need to find du? Is du 1/2(2+e^x)^-1/2??? Thank you
Originally Posted by chocolatelover Is du 1/2(2+e^x)^-1/2??? You forgot the chain rule, try again
Is du 1/2(2+e^x)^-1/2??? You forgot the chain rule OOPs! so du is 1/2(2+e^x)^-1/2(e^x). This won't work for u substitution, right? It wouldn't just be 1/sq. 2art sq. 2+e^x/sq. 2+e^x+c, would it? Thank you
$\displaystyle \int\frac{e^{2x}}{\sqrt{2+e^x}}\,dx$ If $\displaystyle u=\sqrt{2+e^x}\implies du=\frac{e^x}{2\sqrt{2+e^x}}\,dx$, the integral becomes to $\displaystyle \int\frac{e^{2x}}{\sqrt{2+e^x}}\,dx=2\int(u^2-2)\,du$ That's easy, don't you?
$\displaystyle u=\sqrt{2+e^x}\Rightarrow u^2=2+e^x\Rightarrow 2udu=e^xdx$ Substitute and you'll get $\displaystyle \displaystyle2\int(u^2-2)du$
Originally Posted by chocolatelover Hi, Could someone please tell me if this is correct? int. e^2x/sq.2+e^x int. du/a^2+b^2=1/a arctanu/a+c u=2+e^x du=e^xdx int. e^2x/sq. 2+e^x=1/sq.2 arctan 2+e^x/sq. 2+c Thank you Please do us a favor and either learn LaTeX or start using parenthesis! The way you are writing this is confusing. -Dan
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