Question

If \[\int {\dfrac{{{e^x} - 1}}{{{e^x} + 1}}} dx = f\left( x \right) + c\], the find \[f\left( x \right)\].
A. \[2\log \left( {{e^x} + 1} \right)\]
B. \[2\log \left| {{e^{2x}} - 1} \right|\]
C. \[2\log \left| {{e^x} + 1} \right| - x\]
D. \[2\log \left( {{e^{2x}} + 1} \right)\]

Answer 1

Hint: First we will assume \[u = {e^x}\]. Then we calculate its derivatives and find \[dx\] in terms of \[du\] and \[u\]. After that, we will rewrite the integration in terms \[du\] and \[u\]. Then rewrite the expression as \[\dfrac{1}{u}\] and \[\dfrac{1}{{u + 1}}\] , then integrate it by using the formula \[\int {\dfrac{1}{u}} du = \log \left| u \right| + c\]. In the end we will substitute \[u = {e^x}\] to get desired result.

Formula Used:
\[\dfrac{d}{{dx}}{e^x} = {e^x}\]
\[\int {\dfrac{1}{u}} du = \log \left| u \right| + c\]
\[{\log _a}{a^x} = x\]

Complete step by step solution:
Given equation is \[\int {\dfrac{{{e^x} - 1}}{{{e^x} + 1}}} dx = f\left( x \right) + c\]
Assume that \[u = {e^x}\]
Differentiate both sides by \[x\]
\[du = {e^x}dx\]
\[ \Rightarrow dx = \dfrac{{du}}{{{e^x}}}\]
Substitute \[{e^x} = u\]
\[ \Rightarrow dx = \dfrac{{du}}{u}\]
Now putting \[dx = \dfrac{{du}}{u}\] and \[{e^x} = u\] in \[\int {\dfrac{{{e^x} - 1}}{{{e^x} + 1}}} dx\]
\[\int {\dfrac{{{e^x} - 1}}{{{e^x} + 1}}} dx\]
\[ = \int {\dfrac{{u - 1}}{{u + 1}}} \cdot \dfrac{{du}}{u}\]
\[ = \int {\dfrac{{u - 1}}{{u\left( {u + 1} \right)}}} du\]
Add and subtract u with numerator
\[ = \int {\dfrac{{2u - u - 1}}{{u\left( {u + 1} \right)}}} du\]
\[ = \int {\dfrac{{2u - \left( {u + 1} \right)}}{{u\left( {u + 1} \right)}}} du\]
Use the reverse formula \[\dfrac{{a + b}}{c} = \dfrac{a}{c} + \dfrac{b}{c}\]
\[ = \int {\dfrac{{2u}}{{u\left( {u + 1} \right)}}} du + \int {\dfrac{{ - \left( {u + 1} \right)}}{{u\left( {u + 1} \right)}}} du\]
Cancel out common terms
\[ = 2\int {\dfrac{1}{{\left( {u + 1} \right)}}} du - \int {\dfrac{1}{u}} du\]
Apply the formula \[\int {\dfrac{1}{u}} du = \log \left| u \right| + c\]
\[ = 2\log \left| {u + 1} \right| - \log \left| u \right| + c\]
Now substitute \[u = {e^x}\]
\[ = 2\log \left| {{e^x} + 1} \right| - \log \left| {{e^x}} \right| + c\]
Applying the formula \[{\log _a}{a^x} = x\]
\[ = 2\log \left| {{e^x} + 1} \right| - x + c\]
Putting \[\int {\dfrac{{{e^x} - 1}}{{{e^x} + 1}}} dx\]\[ = 2\log \left| {{e^x} + 1} \right| - x + c\] in the equation \[\int {\dfrac{{{e^x} - 1}}{{{e^x} + 1}}} dx = f\left( x \right) + c\] and compute \[f\left( x \right)\].
\[2\log \left| {{e^x} + 1} \right| - x + c = f\left( x \right) + c\]
\[ \Rightarrow f\left( x \right) = 2\log \left| {{e^x} + 1} \right| - x\]

Hence option C is the correct option.

Note: Students often do a common mistake when they use substitution method. They do not change \[dx\] to \[du\]. They solve the integration as \[\int {\dfrac{{u - 1}}{{u + 1}}} du\] and get an incorrect result. The correct integration is \[\int {\dfrac{{u - 1}}{{u\left( {u + 1} \right)}}} du\].