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Question: Integrate with respect to \[x\] , \[\int {\dfrac{{{e^x}}}{{{e^x} - 1}}dx} \] ....

Integrate with respect to xx , ∫exex−1dx\int {\dfrac{{{e^x}}}{{{e^x} - 1}}dx} .

Explanation

Solution

The process of integrating the integrals is known as integration (i.e.) adding up of discrete data. Integration is used to find the function, when the derivatives of that function are given. The basic integration formula which we need to know is ∫xndx=xn+1n+1+c\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c , whereccis integration constant. And other formulae that we will be using are ∫exdx=ex+c\int {{e^x}dx} = {e^x} + c and ∫1xdx=ln⁡∣x∣+c\int {\dfrac{1}{x}dx} = \ln |x| + c , whereccis the integration constant. The basic differentiation formula that we need to know is xn=nxn−1dx{x^n} = n{x^{n - 1}}dx and exdx=ex{e^x}dx = {e^x} .

Complete step by step answer:
The given expression is ∫exex−1dx\int {\dfrac{{{e^x}}}{{{e^x} - 1}}dx} .
Let us make one substitution here. Put ex−1=u{e^x} - 1 = u .
Consider ex−1=u{e^x} - 1 = u , on differentiating this with respect to xx we get,
dudx=ex\dfrac{{du}}{{dx}} = {e^x}
Making some simplification in the above equation we will get,
dx=duexdx = \dfrac{{du}}{{{e^x}}}
On further simplification we get,
dx=e−xdudx = {e^{ - x}}du
On substituting the value of e−x−1{e^{ - x}} - 1 and dxdx this in the given expression we get,
∫exex−1dx=∫1udu\int {\dfrac{{{e^x}}}{{{e^x} - 1}}dx} = \int {\dfrac{1}{u}du}
On integration the above expression with respect to xx we get,
=ln⁡∣u∣+c= \ln |u| + c (Since, ∫1xdx=ln⁡∣x∣+c\int {\dfrac{1}{x}dx} = \ln |x| + c , where c is the integration constant)
Now we have to re-substitute the value of uu , so we get
=ln⁡∣ex−1∣+c= \ln |{e^x} - 1| + c
Thus, we have integrated the given expression that is, ∫exex−1dx=ln⁡∣ex−1∣+c\int {\dfrac{{{e^x}}}{{{e^x} - 1}}dx} = \ln |{e^x} - 1| + c where cc is the integration constant.

Note: There are two types of integrals; they are definite integral and indefinite integral. Definite integrals will have upper limit and lower limit. The integral without limits is called an indefinite integral. The given problem is of the type indefinite integral since it does not have limits. In this problem we have used the substitution method to make the given expression to a simpler form, because we cannot integrate ∫exex−1dx\int {\dfrac{{{e^x}}}{{{e^x} - 1}}dx} directly since it has complexity in their terms.