Simple single variable integral calc

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Homework Statement



e^x dx
1+e^2x

Homework Equations


The Attempt at a Solution



Its the e^x and the e^2x that's tripping me up, i know that e^x is just itself but what is the rule behind e^2x?
 
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u = e^x
du= e^xdx

So then your integral is

u/(1+u^2)du

Does that look like something (think arc(something))
 
\int\frac{e^x}{1+e^{2x}}dx

u=e^x
du=e^xdx

PowerIso meant ...

\int\frac{du}{1+u^2}
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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