$\displaystyle \int \dfrac{e^x+e^{2x}+e^{3x}+e^{4x}}{e^{4x}}\,\mathrm{d}x$

$\displaystyle \int \dfrac{e^x+e^{2x}+e^{3x}+e^{4x}}{e^{4x}}\,\mathrm{d}x= \int \left(e^{-3x}+e^{-2x}+e^{-x}+1\right)\,\mathrm{d}x= \dfrac{e^{-3x}}{-3}+\dfrac{e^{-2x}}{-2}+\dfrac{e^{-x}}{-1}+x+k$ Indeed, for any integral $$\int e^{ax}\,\mathrm{d}x$$ we may apply a simple change of variable $$t=ax$$ $$\mathrm{d}t=a\,\mathrm{d}x$$ so $$\int e^{ax}\,\mathrm{d}x=\dfrac{1}{a}\int e^{ax}a\,\mathrm{d}x=\dfrac{1}{a}\int e^t\,\mathrm{d}t=\dfrac{1}{a}e^t+k=\dfrac{1}{a}e^{ax}+k$$ One could also reason this way: $$\int e^{ax}\,\mathrm{d}x=\int \left(e^a\right)^x\,\mathrm{d}x=\dfrac{1}{\ln\left(e^a\right)}\left(e^a\right)^x+k=\dfrac{1}{a}e^{ax}+k$$ In both ways, $$\int e^{ax}\,\mathrm{d}x=\dfrac{1}{a}e^{ax}+k$$ and also $\displaystyle \int \dfrac{e^x+e^{2x}+e^{3x}+e^{4x}}{e^{4x}}\,\mathrm{d}x= \bbox[#FFECB3,5px]{-\dfrac{e^{-3x}}{3}-\dfrac{e^{-2x}}{2}-e^{-x}+x+k}$

$\displaystyle \int \dfrac{e^{2x}+e^{-2x}}{e^x}\,\mathrm{d}x$

$\displaystyle \int \dfrac{e^{2x}+e^{-2x}}{e^x}\,\mathrm{d}x= \int \left(e^x+e^{-3x}\right)\,\mathrm{d}x= \bbox[#FFECB3,5px]{e^x-\dfrac{e^{-3x}}{3}+k}$

$\displaystyle \int \sin(5x+2)\,\mathrm{d}x$

$\displaystyle \int \sin(5x+2)\,\mathrm{d}x= \dfrac{1}{5}\int \sin(5x+2)5\,\mathrm{d}x \,\overset{\substack{t=5x+2\\ \mathrm{d}t=5\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \dfrac{1}{5}\int \sin t\,\mathrm{d}t= -\dfrac{1}{5}\cos t + k \,\overset{\substack{t=5x+2\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{-\dfrac{1}{5}\cos(5x+2) + k}$