UPSC NDA Math Model Paper 4

Here we have to find the value of ∫

dx

ex−1

Let ex=t and by differentiating eX=t with respect to x we get
⇒exdx=dt or dx=dt∕ex=dt∕t
⇒∫

dx

ez−1

=∫

dt

t(t−1)

Let

1

t(t−1)

=

A

t

+

B

t−1

⇒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
⇒

1

t(t−1)

=

−1

t

+

1

t−1

⇒∫

dt

t(t−1)

=−∫

dt

t

+∫

dt

t−1

As we know that ∫

dx

x

=log|x|+C where C is a constant
⇒∫

dt

t(t−1)

=−log|t| +log|t−1|+C
=log|

t−1

t

|+C
By substituting ex=t in the above equation we get
⇒∫

dx

ez−1

=log|

ex−1

ex

|+C