Question

How do you use partial fractions to find the integral of$\int {\dfrac{{{e^x}}}{{\left( {{e^{2x}} + 1} \right)\left( {{e^x} - 1} \right)}}} dx$?

Answer 1

Hint: Partial fraction is the decomposition of an equation which is of the form$\dfrac{{A(x)}}{{B(x)}}$. It makes it easier to solve the integral as it is represented as the sum or difference of fractions. This method is only used when the fraction is linear in the denominator part.

Complete step by step answer:
The equation is as follows:
$\int {\dfrac{{{e^x}}}{{\left( {{e^{2x}} + 1} \right)\left( {{e^x} - 1} \right)}}} dx$
Let us consider , $y = {e^x}$
Differentiating both sides with respect to $x$
$
  \dfrac{{dy}}{{dx}} = {e^x} \\
  dx = \dfrac{{dy}}{{{e^x}}} \\
 $
Now, we substitute these values to the main equation,
$
  \int {\dfrac{{{e^x}}}{{({y^2} + 1)(y - 1)}}} .\dfrac{{dy}}{{{e^x}}} \\
    \\
 $
Now, we cancel ${e^x}$from the numerator and denominator.
Hence, we get
$\int {\dfrac{1}{{({y^2} + 1)(y - 1)}}} dy$
Now, we use partial fractions
$
  \dfrac{{Ay + B}}{{{y^2} + 1}} + \dfrac{C}{{y - 1}} = \dfrac{1}{{({y^2} + 1)(y - 1)}} \\
  (Ay + B)(y - 1) + C({y^2} + 1) = 1 \\
  (A + C){y^2} + (B - A)y + (C - B) = 1 \\
 $
We now compare the coefficients on both sides,
$
  A + C = 0 \\
  B - A = 0 \\
  C - B = 1 \\
 $
Comparing the above equations we get,
$
  B = A \\
  A = - C \\
 $
Then,
$
  C - B = 1 \\
   - A - A = 1 \\
   - 2A = 1 \\
  A = \dfrac{{ - 1}}{2},B = - \dfrac{1}{2},C = \dfrac{1}{2} \\
 $
Substituting these values to the equation we get,
$
   = \int {\dfrac{{ - \dfrac{1}{2}y - \dfrac{1}{2}}}{{{y^2} + 1}}} + \dfrac{{\dfrac{1}{2}}}{{y - 1}}dy \\
   = - \dfrac{1}{2}\int {\dfrac{{y - 1}}{{{y^2} + 1}}} + \dfrac{1}{{y - 1}}dy \\
   = - \dfrac{1}{2}\int {\dfrac{{y - 1}}{{{y^2} + 1}}} + \dfrac{1}{{y - 1}}dy \\
   = - \dfrac{1}{2}\int {\dfrac{y}{{{y^2} + 1}}} - \dfrac{1}{{{y^2} + 1}} + \dfrac{1}{{y - 1}}dy \\
   = - \dfrac{1}{4}\int {\dfrac{{2y}}{{{y^2} + 1}}} + \dfrac{1}{2}\int {\dfrac{1}{{{y^2} + 1}}} - \dfrac{1}{2}\int {\dfrac{1}{{y - 1}}} dy \\
   = - \dfrac{1}{4}\log ({y^2} + 1) + \dfrac{1}{2}{\tan ^{ - 1}}(y) - \dfrac{1}{2}\log (y - 1) + C \\
 $
Now, substitute$y = {e^x}$, we get
$ = - \dfrac{1}{4}\log ({e^{2x}} + 1) + \dfrac{1}{2}{\tan ^{ - 1}}({e^x}) - \dfrac{1}{2}\log ({e^x} - 1) + C$

Note: If $\dfrac{{A(x)}}{{B(x)}}$ is the equation then the partial fraction is applicable , if and only if, the degree of $A(x)$ is less than that of $B(x)$. If the value is greater, then this method fails and we need to look for some other alternative.