ลองทำครับ
$$ \int \frac{e^x-e^{-x}}{e^x+e^{-x}}dx $$
ให้ $ u =e^x+e^{-x} $
$$ \frac{du}{dx}=e^x-e^{-x} $$
$$ \int \frac{du}{u}=\ln|u|+C $$
$$ \int \frac{e^x-e^{-x}}{e^x+e^{-x}} dx=\ln|e^x+e^{-x}|+C $$
$$\begin{array}{rcl} \int \frac{dx}{1+\sin x}&=&\int\frac{1-\sin x}{1-\sin^2x}\ dx\\
&=&\int \frac{1-\sin x}{\cos^2x}\ dx\\
&=&\int \sec^2x\ dx -\int \tan x\sec x \ dx\\
&=&\tan x - \sec x+C \\
\therefore \int\frac{dx}{1+\sin x} &=&\tan x -\sec x +C\end{array}$$
$$\int \frac{dx}{\sin x} = \int \csc x\ dx=\ln|\csc x - \cot x|+c$$