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