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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals1 Mark
Question
$\int\frac{2}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}\text{ dx}=$
  1. $\frac{-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$
  2. $-\frac{1}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$
  3. $\frac{-1}{(\text{e}^{\text{x}}+1)^2}+\text{C}$
  4. $\frac{1}{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}+\text{C}$
✓

Answer

  1. $\frac{-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$

Solution:

$\text{I}=\int\frac{2}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}\text{ dx}$

$\text{I}=\int\frac{2\text{e}^{2\text{x}}}{(\text{e}^{2\text{x}}+1)^2}\text{ dx}$

Put $\text{t}=\text{e}^{2\text{x}}+1$

$\text{dt}=2\text{e}^{2\text{x}}\text{ dx}$

$\text{I}=\int\frac{\text{dt}}{\text{t}^2}$

$\text{I}=\frac{-1}{\text{t}}+\text{C}$

$\text{I}=\frac{-1}{\text{e}^{2\text{x}}+1}+\text{C}$

$\text{I}=\frac{-\frac{1}{\text{e}^{\text{x}}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$

$\text{I}=\frac{-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$

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