Evaluate: {dx}{e^{x}-1}

Correct Answer is : log⁡∣ex−1ex∣+C ≤ft| {e^{x}-1}{e^{x}} | + Clogexex−1+C

- ≤ft| {e^{x}-1}{e^{x}} | + C

Correct Answer is : log⁡∣ex−1ex∣+C ≤ft| {e^{x}-1}{e^{x}} | + Clogexex−1+C

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UPSC NDA Math Model Paper 4

Question : 108 of 120

Marks: +1, -0

Evaluate:

\int\limits \frac{dx}{e^{x}-1}

Solution:

Here we have to find the value of

\int\limits \frac{dx}{e^{x}-1}

Let

e^{x}=t

and by differentiating

e^{X}=t

with respect to

x

we get

\Rightarrow e^{x} dx = dt

dx = \frac{dt}{e^{x}} = \frac{dt}{t}

\Rightarrow \int\limits \frac{dx}{e^{z}-1} = \int\limits \frac{dt}{t(t-1)}

Let

\frac{1}{t(t-1)} = \frac{A}{t} + \frac{B}{t-1}

\Rightarrow 1 = A(t-1) + Bt

......(1)

By putting t = 0 on both the sides of (1) we get A = - 1

By putting t = 1 on both the sides of (1) we get B = 1

\Rightarrow \frac{1}{t(t-1)} = \frac{-1}{t} + \frac{1}{t-1}

\Rightarrow \int\limits \frac{dt}{t(t-1)} = -\int\limits \frac{dt}{t} + \int\limits \frac{dt}{t-1}

As we know that

\int\limits \frac{dx}{x} = \log |x| + C

where

C

is a constant

\Rightarrow \int\limits \frac{dt}{t(t-1)} = -\log |t|

+\log |t-1| + C

= \log \left| \frac{t-1}{t} \right| + C

By substituting

e^{x}=t

in the above equation we get

\Rightarrow \int\limits \frac{dx}{e^{z}-1} = \log \left| \frac{e^{x}-1}{e^{x}} \right| + C

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