-01d x-ex-e-x isA- -4-tan -1 e B- tan -1 e -4C- tan -1 e -4D- tan -1 e

The correct option is B tan^ 1(e) π/4 nbsp; nbsp; nbsp;I= nbsp;∫_0^1dx/e^x+e^ x=∫_0^1e^xdx/e^2x+1 nbsp; Put e^x=t ⇒ e^x dx =dt Then I=∫_1^edt/1+t^2 ...

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Byju's Answer

Standard XII

Mathematics

Derivative from First Principle

∫01d x/ex+e-x...

Question

$1 \int 0 \frac{d x}{e^{x} + e^{- x}}$ is

\frac{π}{4} - {tan}^{- 1} (e)

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{tan}^{- 1} (e) - \frac{π}{4}

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{tan}^{- 1} (e) + \frac{π}{4}

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{tan}^{- 1} (e)

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Solution

The correct option is B ${tan}^{- 1} (e) - \frac{π}{4}$
$I = 1 \int 0 \frac{d x}{e^{x} + e^{- x}} = 1 \int 0 \frac{e^{x} d x}{e^{2 x} + 1}$
$Put e^{x} = t \Rightarrow e^{x} d x = d t$
$Then I = e \int 1 \frac{d t}{1 + t^{2}}$
$= [{tan}^{- 1} t]_{1}^{e} = {tan}^{- 1} (e) - \frac{π}{4}$

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Similar questions

Choose the correct answer.

$\int \frac{1}{e^{x} + e^{- x}} d x i s e q u a l t o (a) t a n^{- 1} e^{x} + C (b) t a n^{- 1} e^{- x} + C (c) l o g (e^{x} - e^{- x}) + C (d) l o g (e^{x} + e^{- x}) + C$

$t a n^{- 1} (\frac{x}{y}) - t a n^{- 1} (\frac{x - y}{x + y})$ is equal to
a) $\frac{π}{2}$
b) $\frac{π}{3}$
c) $\frac{π}{4}$
d) $\frac{- 3 π}{4}$

Q. If

y = s i n^{- 1} x + t a n^{- 1} x + s e c^{- 1} x

, then set of values of y is

$t a n^{- 1} \frac{a}{b} - t a n^{- 1} (\frac{a - b}{a + b})$ =

\sum_{r = 1}^{n} t a n^{- 1} (\frac{2^{r - 1}}{1 + 2^{2 r - 1}})

is equal to

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1∫0dxex+e−x is

The correct option is B tan−1(e)−π4 I=1∫0dxex+e−x=1∫0exdxe2x+1 Put ex=t⇒exdx=dt Then I=e∫1dt1+t2 =[tan−1t]e1=tan−1(e)−π4

$1 \int 0 \frac{d x}{e^{x} + e^{- x}}$ is

The correct option is B ${tan}^{- 1} (e) - \frac{π}{4}$
$I = 1 \int 0 \frac{d x}{e^{x} + e^{- x}} = 1 \int 0 \frac{e^{x} d x}{e^{2 x} + 1}$
$Put e^{x} = t \Rightarrow e^{x} d x = d t$
$Then I = e \int 1 \frac{d t}{1 + t^{2}}$
$= [{tan}^{- 1} t]_{1}^{e} = {tan}^{- 1} (e) - \frac{π}{4}$