# Evaluate int(9e^(9t))/(e^(18t)+11e^(9t)+18)dt?

May 9, 2018

See below

#### Explanation:

Intuition make that we try the change $z = {e}^{9 t}$ with this we have

$\mathrm{dz} = 9 {e}^{9 t} \mathrm{dt}$ and integral changes to

$\int \frac{\mathrm{dz}}{{z}^{2} + 11 z + 18} = I$

Now factorize $\frac{1}{{z}^{2} + 11 z + 18} = \frac{1}{\left(z + 2\right) \left(z + 9\right)} = \frac{A}{z + 2} + \frac{B}{z + 9}$

We found $A = \frac{1}{7}$ and $B = - \frac{1}{7}$

And then $I = \frac{1}{7} \int \frac{\mathrm{dz}}{z + 2} - \frac{1}{7} \int \frac{\mathrm{dz}}{z + 9} = \frac{1}{7} \ln \left(z + 2\right) - \frac{1}{7} \ln \left(z + 9\right) = \frac{1}{7} \ln \left({e}^{9 t} + 2\right) - \frac{1}{7} \ln \left({e}^{9 t} + 9\right) + C$

We can re-arrange applying logarithm laws

$I = \frac{1}{7} \ln \left(\frac{{e}^{9 t} + 2}{{e}^{9 t} + 9}\right)$

May 9, 2018

$\frac{1}{7} \ln | \left(\frac{{e}^{9 t} + 2}{{e}^{9 t} + 2}\right) | + C .$

#### Explanation:

Observe that, ${e}^{18 t} + 11 {e}^{9 t} + 18 = \left({e}^{9 t} + 9\right) \left({e}^{9 t} + 2\right)$.

So, let, ${e}^{9 t} = x . \therefore {e}^{9 t} \cdot 9 \mathrm{dt} = \mathrm{dx}$.

$\therefore I = \int \frac{9 {e}^{9 t}}{{e}^{18 t} + 11 {e}^{9 t} + 18} \mathrm{dt}$,

$= \int \frac{1}{\left(x + 9\right) \left(x + 2\right)} \mathrm{dx}$,

$= \frac{1}{7} \int \frac{\left(x + 9\right) - \left(x + 2\right)}{\left(x + 9\right) \left(x + 2\right)} \mathrm{dx}$,

$= \frac{1}{7} \int \left\{\frac{x + 9}{\left(x + 9\right) \left(x + 2\right)} - \frac{x + 2}{\left(x + 9\right) \left(x + 2\right)}\right\} \mathrm{dx}$,

$= \frac{1}{7} \int \left\{\frac{1}{x + 2} - \frac{1}{x + 9}\right\} \mathrm{dx}$,

$= \frac{1}{7} \left\{\ln | \left(x + 2\right) - \ln | \left(x + 9\right) |\right\}$,

$= \frac{1}{7} \ln | \frac{x + 2}{x + 9} |$.

$\Rightarrow I = \frac{1}{7} \ln | \left(\frac{{e}^{9 t} + 2}{{e}^{9 t} + 2}\right) | + C .$

Enjoy Maths.!