Choose the correct answer-a tan -1 e - x - Cllb tan -1 e - x - C -c log- e - x - e - x - Clld log- e - x - e - x - C l

I=∫1/e^x+e^ x=∫1/e^x+1/e^xdx=∫e^x/(e^x)^2+1dx Put e^x=t⇒ e^xdx=dt (∵ ∫ dx/1+x^2 =tan^ 1x ) ∴ I=∫ e^x/t^2+1dt/e^x=tan^ 1(t)+C.=tan^ 1e^x+C. Hence, opt ...

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Byju's Answer

Standard XII

Mathematics

Definite Integral as Limit of Sum

Choose the co...

Question

Choose the correct answer.

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

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Solution

$I = \int \frac{1}{e^{x} + e^{- x}} = \int \frac{1}{e^{x} + \frac{1}{e^{x}}} d x = \int \frac{e^{x}}{(e^{x})^{2} + 1} d x P u t e^{x} = t \Rightarrow e^{x} d x = d t (∵ \int \frac{d x}{1 + x^{2}} = t a n^{- 1} x) ∴ I = \int \frac{e^{x}}{t^{2} + 1} \frac{d t}{e^{x}} = t a n^{- 1} (t) + C . = t a n^{- 1} e^{x} + C .$
Hence, option (a) is correct.

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Similar questions

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

Q. When the tangent to the curve y = x log x is parallel to the chord joining the points (1, 0) and (e, e), the value of x is

(a) e¹^/1−e

(b) e^{(e−1)(2e−1)}

(c)

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

(d)

\frac{e - 1}{e}

Q. Match the following.
I.

\int e^{x} (sin x + cos x) d x =

e^{x} tan x + c

II.

\int e^{x} (cos x - sin x) d x =

e^{x} log sec x + c

III.

\int e^{x} (tan x + {sec}^{2} x) d x =

e^{x} sin x + c

IV.

\int e^{x} (tan x + log sec x) d x =

e^{x} cos x + c

Q. Integration of

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

with respect to log_e x is
(a)

\frac{\tan^{- 1} (\log_{e} x)}{x} + C

(b)

\tan^{- 1} (\log_{e} x) + C

(c)

\frac{\tan^{- 1} x}{x} + C

(d) none of these

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

(a) 2 log_e cos (xe^x) + C

(b) sec (xe^x) + C

(c) tan (xe^x) + C

(d) tan (x + e^x) + C

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Choose the correct answer. ∫1ex+e−x dx is equal to(a) tan−1ex+C(b) tan−1e−x+C (c) log (ex−e−x)+C(d) log (ex+e−x)+C

I=∫1ex+e−x=∫1ex+1exdx=∫ex(ex)2+1dxPut ex=t⇒ exdx=dt (∵ ∫ dx1+x2=tan−1x)∴ I=∫ ext2+1dtex=tan−1(t)+C.=tan−1ex+C. Hence, option (a) is correct.

Choose the correct answer.

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