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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS1 Mark
Question
Integrate the functions in Exercises:
$\frac{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}$
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Answer

$\text{Let I}=\int\frac{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}} \text{ dx}=\frac{1}{2}\int\frac{\text 2\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)}{\text {e}^{\text{2x}}+\text{e}^{-\text{2x}}}\text{ dx} \ \ \ \ \ \ ...\text{(i)} $
Putting$\text{ e}^{\text{2x}}+\text{e}^{-2\text{x}}=t\ \ \ \Rightarrow \ \ \ \ \text{ e}^{\text{2x}}\frac{\text{d}}{\text{dx }} 2\text{x} +\text{e}^{-2\text{x}}\frac{\text{d}}{\text{dx}}(-2\text{x})=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{2\text{x}}.2-2\text{e}^{-2\text{x}}=\frac{\text{dt}}{\text{dx}} \ \ \ \ \Rightarrow\ \ \ \ \ 2\big(\text{e}^{\text{2x}}-\text{e}^{-\text{2x}}\big)\text{ dx}=\text{dt} $
$\therefore \ \ \ \ \ $From eq. (i),    $\text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}} $
$=\frac{1}{2}\log\begin{vmatrix}t\end{vmatrix}+\text{c}=\frac{1}{2}\log\begin{vmatrix}\text{e}^\text {2x}+\text{e}^{-\text{2x}}\end{vmatrix}+\text{c} $
$=\frac{1}{2}\log\big(\text{e}^\text{2x}+\text{e}^{-\text{2x}}\big)+\text{c} $

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