Question:
$\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}$
Solution:
Let $e^{2 x}+e^{-2 x}=t$
$\therefore\left(2 e^{2 x}-2 e^{-2 x}\right) d x=d t$
$\Rightarrow 2\left(e^{2 x}-e^{-2 x}\right) d x=d t$
$\Rightarrow \int\left(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\right) d x=\int \frac{d t}{2 t}$
$=\frac{1}{2} \int \frac{1}{t} d t$
$=\frac{1}{2} \log |t|+\mathrm{C}$
$=\frac{1}{2} \log \left|e^{2 x}+e^{-2 x}\right|+\mathrm{C}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.