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Question: Find the value of by integrating \[\int {\dfrac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} dx\]?...

Find the value of by integrating ∫e2x−1e2x+1dx\int {\dfrac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} dx?

Explanation

Solution

The above question is based on the concept of integration. Since it is an indefinite integral which has no upper and lower limits, we can apply integration properties by substituting the expression and we can find the antiderivative of the above expression.

Complete step by step solution:
Integration is a way of finding the antiderivative of any function. It is the inverse of differentiation. It denotes the summation of discrete data. Calculation of small problems is an easy task but for adding big problems which include higher limits, integration method is used. The above given expression is an indefinite integral which means there are no upper or lower limits given.
The above expression after integrating should be in the below given form.
∫f(x)=F(x)+C\int {f\left( x \right) = F\left( x \right) + C}
where C is constant and f(x) is function.
So, the given expression is ∫e2x−1e2x+1dx\int {\dfrac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} dx
So, we need to first bring it in terms of e we get,
⇒∫(e2x−1)×e−x(e2x+1)×e−x\Rightarrow \int {\dfrac{{\left( {{e^{2x}} - 1} \right) \times {e^{ - x}}}}{{\left( {{e^{2x}} + 1} \right) \times {e^{ - x}}}}}
Here we have multiplied with e−x{e^{ - x}}.And now further multiplying on opening the parenthesis.
⇒∫ex−e−xex+e−xdx\Rightarrow \int {\dfrac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}dx}
So, now let (ex+e−x)=t\left( {{e^x} + {e^{ - x}}} \right) = t
By taking derivative of it on both the sides we get,

⇒(ex+(−1)e−x)dx=dt ⇒ex−e−x=dtdx ⇒dx=dtex−e−x \Rightarrow \left( {{e^x} + \left( { - 1} \right){e^{ - x}}} \right)dx = dt \\\ \Rightarrow {e^x} - {e^{ - x}} = \dfrac{{dt}}{{dx}} \\\ \Rightarrow dx = \dfrac{{dt}}{{{e^x} - {e^{ - x}}}} \\\

Now by substituting ex+e−x=t{e^x} + {e^{ - x}} = t and dx=dtex+e−xdx = \dfrac{{dt}}{{{e^x} + {e^{ - x}}}} in the above equation,

⇒∫ex−e−xt×dtex−e−x ⇒∫1tdt \Rightarrow \int {\dfrac{{{e^x} - {e^{ - x}}}}{t}} \times \dfrac{{dt}}{{{e^x} - {e^{ - x}}}} \\\ \Rightarrow \int {\dfrac{1}{t}dt} \\\

Now by integrating the expression we get,
⇒log⁡t=log⁡(ex+e−x)\Rightarrow \log t = \log \left( {{e^x} + {e^{ - x}}} \right)
Therefore, we get the above solution.

Note: An important thing to note is that when we substitute the expression with the variable t and then it is differentiated where the power -x is differentiated in such a way that the coefficient is multiplied with the expression e along with the negative sign.