What trick is used to integrate this?

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Homework Statement



\int e^x \frac{1 + sin(x)}{1 + cos(x)}\;dx


The Attempt at a Solution



Apparently there is a trick involving e^x integrals like this

\int e^x[f(x) + f'(x)] dx = e^x f(x) + C

Now my question is not how to compute the integral above but where did this \int e^x[f(x) + f'(x)] dx = e^x f(x) + C identity came from?? How does one know this exists?
 
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Take the derivative of the result. That may help you see how this works.
 
Yeah, you get this result with integration "by parts", which is the name for using the product rule to rewrite integrals (hopefully in a form you can recognize and solve). If you take a second-semester single-variable calculus class you'll learn about this.
 
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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