Question

$\int_{}^{}{\frac{\mathbf{dx}}{\mathbf{(}\mathbf{e}^{\mathbf{x}}\mathbf{+}\mathbf{e}^{\mathbf{-}\mathbf{x}}\mathbf{)}^{\mathbf{2}}}\mathbf{=}}$

• A. $\tan hx + c$
• B. $\frac{1}{2}\tan hx + c$
• C. $\frac{1}{4}\tan hx + c$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{4}\tan hx + c$

Answer 2

$\int_{}^{}\frac{dx}{(e^{x} + e^{- x})^{2}} = \frac{1}{4}\int_{}^{}\frac{4.dx}{(e^{x} + e^{- x})^{2}} = \frac{1}{4}\int_{}^{}{\sec h^{2}x}dx = \frac{1}{4}\tan ⥂ hx + c.$