Question
$\int \frac{5 e ^x}{\left( e ^x+1\right)\left( e ^{2 x}+9\right)} d x$

Answer

Let $i =\int \frac{5 e ^x}{\left( e ^x+1\right)\left( e ^{2 x}+9\right)} d x$
Put $e ^{ x }= t$
$ \therefore e ^{ x } dx = dt$
$\therefore I =\int \frac{5 dt }{( t +1)\left( t ^2+9\right)} $
Let $\frac{5}{(t+1)\left(t^2+9\right)}$
$ =\frac{A}{t+1}+\frac{B t+C}{t^2+9}$
$\therefore 5=A\left(t^2+9\right)+(B t+C)(t+1)\ldots(i) $
Putting $t=-1$ in (i), we get
$ 5= A \left[(-1)^2+9\right]$
$\therefore 5=10 A$
$\therefore A =\frac{1}{2} $
Putting $t=0$ in (i), we get
$ 5=A(0+9)+(0+C)(0+1)$
$\therefore 5=9 A+C$
$\therefore 5=9\left(\frac{1}{2}\right)+C $
$\therefore C=\frac{1}{2}$
Putting $t=1$ in (i), we get
$ 5= A \left(1^2+9\right)+( B + C )(1+1)$
$\therefore 5=10 A +2 B +2 C$
$\therefore 5=10\left(\frac{1}{2}\right)+2 B +2\left(\frac{1}{2}\right)$
$\therefore-1=2 B$
$\therefore B =-\frac{1}{2}$
$\therefore \frac{5}{( t +1)\left( t ^2+9\right)}=\frac{\frac{1}{2}}{ t +1}+\frac{\frac{1}{2} t +\frac{1}{2}}{ t ^2+9}$
$\therefore I =\int\left(\frac{\frac{1}{2}}{ t +1}+\frac{\frac{-1}{2} t +\frac{1}{2}}{ t ^2+9}\right) dt$
$=\frac{1}{2}\left[\int \frac{1}{ t +1} dt -\int \frac{ t }{ t ^2+9} dt +\int \frac{1}{ t ^2+9} dt \right]$
$=\frac{1}{2}\left[\int \frac{1}{ t +1} dt -\frac{1}{2} \int \frac{2 t }{ t ^2+9} dt +\int \frac{1}{ t ^2+3^2} dt \right]$
$=\frac{1}{2}\left[\log | t + t |-\frac{1}{2} \log \left| t ^2+9\right|+\frac{1}{3} tan ^{-1}\left(\frac{ t }{3}\right)\right]+ c$
$\therefore I =\frac{1}{2} \log \left| e ^x+1\right|-\frac{1}{4} \log \left| e ^{2 x}+9\right|+\frac{1}{6} tan ^{-1}\left(\frac{ e ^x}{3}\right)+ c $

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