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Question: Evaluate \(\int {\dfrac{{{e^x}}}{{{e^x} + 1}}dx} \) ?...

Evaluate ∫exex+1dx\int {\dfrac{{{e^x}}}{{{e^x} + 1}}dx} ?

Explanation

Solution

First of all, we will add and subtract a number in the numerator, so that they get separated. After that, we will let the denominator as any variable (say t) and will convert dx into dt. Then after solving further we will re-put the values that we have left.

Complete Step by Step Solution:
Let us consider ∫exex+1dx\int {\dfrac{{{e^x}}}{{{e^x} + 1}}dx} as y
⇒y=∫exex+1dx\Rightarrow y = \int {\dfrac{{{e^x}}}{{{e^x} + 1}}dx}
Now, we will add and subtract 1 in the numerator
⇒y=∫ex+1−1ex+1dx\Rightarrow y = \int {\dfrac{{{e^x} + 1 - 1}}{{{e^x} + 1}}dx}
Now, we will separate it into two parts (first one will be the ∫ex+1ex+1dx\int {\dfrac{{{e^x} + 1}}{{{e^x} + 1}}dx} and the other one is ∫−1ex+1dx\int {\dfrac{{ - 1}}{{{e^x} + 1}}dx} )
⇒y=∫ex+1ex+1dx+∫−1ex+1dx\Rightarrow y = \int {\dfrac{{{e^x} + 1}}{{{e^x} + 1}}dx + \int {\dfrac{{ - 1}}{{{e^x} + 1}}} } dx
We can also write it as
⇒y=∫ex+1ex+1dx−∫1ex+1dx\Rightarrow y = \int {\dfrac{{{e^x} + 1}}{{{e^x} + 1}}dx} - \int {\dfrac{1}{{{e^x} + 1}}dx}
⇒y=∫1dx−∫1ex+1dx\Rightarrow y = \int {1dx} - \int {\dfrac{1}{{{e^x} + 1}}dx}
As we know that ∫1dx=x\int {1dx = x}
Therefore, ⇒y=x−∫1ex+1dx\Rightarrow y = x - \int {\dfrac{1}{{{e^x} + 1}}dx}
Let us consider −∫1ex+1dx- \int {\dfrac{1}{{{e^x} + 1}}dx} as I
⇒y=x+I\Rightarrow y = x + I
And I=−∫1ex+1dxI = - \int {\dfrac{1}{{{e^x} + 1}}dx} ……(i)
Let ex+1=t{e^x} + 1 = t ……(ii)
Differentiating both sides of the above equation with respect to t
⇒d(ex)dt+d(1)dt=d(t)dt\Rightarrow \dfrac{{d\left( {{e^x}} \right)}}{{dt}} + \dfrac{{d\left( 1 \right)}}{{dt}} = \dfrac{{d\left( t \right)}}{{dt}}
On further simplification,
⇒ex(dxdt)+0=1\Rightarrow {e^x}\left( {\dfrac{{dx}}{{dt}}} \right) + 0 = 1
⇒ex(dx)=dt\Rightarrow {e^x}\left( {dx} \right) = dt
⇒dx=dtex\Rightarrow dx = \dfrac{{dt}}{{{e^x}}} ……(iii)
Now, by putting the value of dx and ex+1{e^x} + 1 from (ii) and (iii) in (i), we get
⇒I=−∫1t(t−1)dt\Rightarrow I = - \int {\dfrac{1}{{t(t - 1)}}dt}
Now, adding and subtracting t in the numerator
⇒I=−∫1+t−tt(t−1)dt\Rightarrow I = - \int {\dfrac{{1 + t - t}}{{t(t - 1)}}dt}
We can rewrite the above equation as
⇒I=−∫t−(t−1)t(t−1)dt\Rightarrow I = - \int {\dfrac{{t - \left( {t - 1} \right)}}{{t(t - 1)}}dt}
Now, we will separate the above equation into two parts (first one will be the −∫tt(t−1)dt- \int {\dfrac{t}{{t\left( {t - 1} \right)}}dt} and the other one is −∫−(t−1)t(t−1)dt- \int {\dfrac{{ - \left( {t - 1} \right)}}{{t\left( {t - 1} \right)}}dt} )
⇒I=−∫tt(t−1)dt−∫−(t−1)t(t−1)dt\Rightarrow I = - \int {\dfrac{t}{{t\left( {t - 1} \right)}}dt - \int {\dfrac{{ - \left( {t - 1} \right)}}{{t\left( {t - 1} \right)}}dt} }
⇒I=−∫1t−1dt−∫−1tdt\Rightarrow I = - \int {\dfrac{1}{{t - 1}}dt - \int {\dfrac{{ - 1}}{t}dt} }
As we know that ∫1xdx=ln⁡(x)\int {\dfrac{1}{x}dx = \ln \left( x \right)}
Hence I=−ln⁡(t−1)+ln⁡(t)I = - \ln \left( {t - 1} \right) + \ln \left( t \right)
Now, putting the value of t from (ii)
⇒I=−ln⁡(ex+1−1)+ln⁡(ex+1)\Rightarrow I = - \ln \left( {{e^x} + 1 - 1} \right) + \ln \left( {{e^x} + 1} \right)
We can also rewrite the above equation as
⇒I=ln⁡(ex+1)−ln⁡(ex)\Rightarrow I = \ln \left( {{e^x} + 1} \right) - \ln \left( {{e^x}} \right)
As y=x+Iy = x + I
As we know ln⁡(ex)=x\ln \left( {{e^x}} \right) = x

Therefore, y=x+ln⁡(ex+1)−xy = x + \ln ({e^x} + 1) - x
⇒y=ln⁡(ex+1)\Rightarrow y = \ln ({e^x} + 1)

Note:
While doing these types of problems, strictly take care of dx and dt. When you let a variable (say x) to another variable (say t) then take care that you do not forget to change the dx into dt. And in the last, do not forget to put the given variables. Do not leave the answer in those variables which you have taken (let).