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Question: Integrate \(\int {\dfrac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} dx\)....

Integrate ∫e2x−1e2x+1dx\int {\dfrac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} dx.

Explanation

Solution

Here to solve this, we do not have the proper formula. So we will try to convert it into the form which is solvable. In relation to ex{e^x} we know that its integration with respect to xx is ex{e^x} itself. So we need to convert the whole given term into a term which we can integrate easily. So first of all we will divide the denominator and numerator by ex{e^x} then we will get ∫ex−e−xex+e−xdx\int {\dfrac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} dx.
Now if we will let the denominator to be any variable then we will get its differentiation as the numerator. Hence now we will be able to solve it completely.

Complete step by step solution:
Here we are given the term to integrate which is ∫e2x−1e2x+1dx\int {\dfrac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} dx
Here to solve this, we do not have the proper formula. So we will try to convert it into the form which is solvable. In relation to ex{e^x} we know that its integration with respect to xx is ex{e^x} itself. So we need to convert the whole given term into a term which we can integrate easily.
We have ∫e2x−1e2x+1dx\int {\dfrac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} dx
To convert into simpler form let us divide the numerator and denominator by ex{e^x}
Now we get
∫e2x−1exe2x+1exdx\int {\dfrac{{\dfrac{{{e^{2x}} - 1}}{{{e^x}}}}}{{\dfrac{{{e^{2x}} + 1}}{{{e^x}}}}}} dx
Now we can write it as
∫ex−e−xex+e−xdx\int {\dfrac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} dx
Now let the denominator which is ex+e−x{e^x} + {e^{ - x}} as a variable tt
So ex+e−x=t{e^x} + {e^{ - x}} = t
Differentiating both sides we get
(ex−e−x)dx=dt({e^x} - {e^{ - x}})dx = dt
We know ddxex=ex\dfrac{d}{{dx}}{e^x} = {e^x}
Now putting these values in the above part we get
∫dtt\int {\dfrac{{dt}}{t}}
Now we know that ∫dxx=log⁡x+c\int {\dfrac{{dx}}{x}} = \log x + c where cc is any constant.
So we get that ∫dtt=log⁡t+c\int {\dfrac{{dt}}{t}} = \log t + c
Now we can write the values of tt in the above equation and also the value of dtdt
Therefore we get ∫ex−e−xex+e−xdx=log⁡(ex+e−x)+c\int {\dfrac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} dx = \log ({e^x} + {e^{ - x}}) + c

Now multiplying the denominator and denominator by ex{e^x} we get
∫e2x−1e2x+1dx\int {\dfrac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} dx =log⁡(ex+e−x)+c = \log ({e^x} + {e^{ - x}}) + c

Note:
Here a student must take care that while integrating any term we need to first convert it into the form which is solvable easily. This will come by practicing more and more questions. We need to take care while substituting the correct value and their respective derivative.