Q.Choose the correct option: $\int \frac{e^{2x}+e^{-2x}}{e^x}dx =$
(A) $e^x-\frac{1}{3e^{3x}}+c$ (B) $e^x+\frac{1}{3e^{3x}}+c$ (C) $e^{-x}+\frac{1}{3e^{3x}}+c$ (D) $e^{-x}-\frac{1}{3e^{3x}}+c$

Every basic differentiation rule can be read backwards to give a ready-made integration formula. Reversing the power rule gives $\int x^n\,dx=\frac{x^{n+1}}{n+1}+c$ ($n\neq-1$); reversing the exponential rules gives $\int a^x\,dx=\frac{a^x}{\log a}+c$ and $\int e^x\,dx=e^x+c$; reversing the trigonometric derivative rules gives $\int\cos x\,dx=\sin x+c$, $\int\sin x\,dx=-\cos x+c$, $\int\sec^2x\,dx=\tan x+c$, $\int\sec x\tan x\,dx=\sec x+c$, $\int\csc x\cot x\,dx=-\csc x+c$, $\int\csc^2x\,dx=-\cot x+c$; and reversing the logarithm rule gives $\int\frac1x\,dx=\log|x|+c$. Every one of these has a matching "linear-argument" version, replacing $x$ by $ax+b$ …