The correct option is A 12(e^2x + 1) + c ∫1(e^x+e^ x)^2dx =∫e^2x(e^2x+1)^2dx substitute u=e^2x+1⇒ 2dx=e^ 2xdu =12∫1u^2 du = 12u ...

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Byju's Answer

Standard XII

Mathematics

Integration Using Substitution

∫1ex + e-x 2 ...

Question

$\int \frac{1}{(e^{x} + e^{- x})^{2}} d x =$

- \frac{1}{2 (e^{2 x} + 1)} + c

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\frac{1}{2 (e^{2 x} + 1)} + c

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- \frac{1}{e^{2 x} + 1}

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None of these

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Solution

The correct option is A $- \frac{1}{2 (e^{2 x} + 1)} + c$
$\int \frac{1}{(e^{x} + e^{- x})^{2}} d x$

$= \int \frac{e^{2 x}}{(e^{2 x} + 1)^{2}} d x$

substitute $u = e^{2 x} + 1 \Rightarrow 2 d x = e^{- 2 x} d u$

$= \frac{1}{2} \int \frac{1}{u^{2}} d u$

$= \frac{- 1}{2 u}$

$= - \frac{1}{2 (e^{2 x} + 1)} + C$

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∫1(ex+e−x)2dx=

The correct option is A −12(e2x+1)+c∫1(ex+e−x)2dx=∫e2x(e2x+1)2dxsubstitute u=e2x+1⇒2dx=e−2xdu=12∫1u2du=−12u=−12(e2x+1)+C

$\int \frac{1}{(e^{x} + e^{- x})^{2}} d x =$