Question

Make a substitution to express the integrand as a rational function and then evaluate the integral. (Use C for the constant of integration.) ∫ 5e^2x /(e^2x + 13e^x + 36). dx

Answer 1

Answer:

$I=\frac{5}{2}\ln \left ( e^{2x}+13e^x+36 \right )-\frac{13}{2}\left [ \ln (e^x+4)-\ln (e^x+9) \right ]+C$

Step-by-step explanation:

To find:$\int \frac{5e^{2x}}{e^{2x}+13e^x+36}$

Solution:

Let $I =\int \frac{5e^{2x}}{e^{2x}+13e^x+36}$

Take $e^x=t\Rightarrow e^x\,dx=dt$

So,

$I=\int \frac{5t\,dt}{t^2+13t+36}\\=5\int \frac{t\,dt}{t^2+13t+36}\\=\frac{5}{2}\int \frac{2t+13-13}{t^2+13t+36}\,dt\\=\frac{5}{2}\int \frac{2t+13}{t^2+13t+36}\,dt-\frac{65}{2}\int \frac{dt}{t^2+13t+36}$

Here,

$\frac{5}{2}\int \frac{2t+13}{t^2+13t+36}\,dt=\frac{5}{2}\ln \left ( t^2+13t+36 \right )\,\,\left \{ \because \int \frac{f'(x)}{f(x)}\,dx=\ln f(x) \right \}$

$-\frac{65}{2}\int \frac{dt}{t^2+13t+36}=-\frac{65}{2}\int \frac{dt}{t^2+9t+4t+36}\\=-\frac{65}{2}\int \frac{dt}{(t+4)(t+9)}\\=\frac{-65}{2}\left ( \frac{1}{5} \right )\int \frac{1}{t+4}-\frac{1}{t+9}\,dt\\=\frac{-13}{2}\left [ \ln (t+4)-\ln (t+9) \right ]\,\,\left \{ \because \int \frac{dt}{t}=\ln t \right \}$

So,

$I=\frac{5}{2}\int \frac{2t+13}{t^2+13t+36}\,dt-\frac{65}{2}\int \frac{dt}{t^2+13t+36}\\=\frac{5}{2}\ln \left ( t^2+13t+36 \right )-\frac{13}{2}\left [ \ln (t+4)-\ln (t+9) \right ]+C$

C is a constant of integration.

Put $t=e^x$

$I=\frac{5}{2}\ln \left ( e^{2x}+13e^x+36 \right )-\frac{13}{2}\left [ \ln (e^x+4)-\ln (e^x+9) \right ]+C$