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NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If ∫ dx e^ x - 1 = 2tan^ - 1 (f(x)) + C, (where x > 0 and C is the...

If

∫ \dfrac{dx}{e^{x} - 1} = 2tan^{- 1}\left(f\left(x\right)\right) + C,

(where

x > 0

and

C

is the constant of integration) then the range of

f \left(x\right) is

Options

A $\left(0, \infty\right)$
B $\left[0, \infty\right)$
C $\left[1, \infty\right)$
D $\left(1, \infty\right)$

Correct answer

A. $\left(0, \infty\right)$

Step-by-step solution

Le,t

e^{x} - 1 = t^{2}

⇒ e^{x} ⋅ dx = 2tdt

∴ Ι = ∫ \dfrac{2tdt}{t\left(e^{x}\right)}

= ∫ \dfrac{2dt}{t^{2} + 1}

= 2tan^{- 1}t + C

= 2tan^{- 1}\left(\sqrt{e^{x} - 1}\right) + C

∴ f \left(x\right) = \sqrt{e^{x} - 1}

For

x \in \left(0, \infty\right)

; e^{x} \in \left(1, \infty\right)

<br />

⇒ \sqrt{e^{x} - 1} \in \left(0, \infty\right)

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