Integration exercise: $ \int \frac{e^{5x}}{ (e^{2x} - e^x - 20) }dx$

Question

I have trouble integrating:

$$ \int \frac{e^{5x}}{e^{2x} - e^x - 20} dx$$

With $t=e^x$, I've rewritten it as:

$$\int \frac{t^5}{t^2 - t - 20} \frac{1}{t} dt$$

Then I tried integration by parts, but I am not any closer to the solution.

Answer 1

$$\int \frac{t^5}{t^2 - t - 20} \frac{1}{t} dt$$ $$\int \frac{t^4}{t^2 - t - 20} dt$$
try to make proper fraction $$\int (t^2+t+21)+\frac{41t+420}{t^2 - t - 20} dt$$ $$\int (t^2+t+21)\;dt+\int \frac{41t+420}{t^2 - t - 20} dt$$ first integration is basic and for second use partial fraction $$\dfrac{t^3}{3}+\dfrac{t^2}{2}+21t+\int \dfrac {625}{9(t-5)}-\dfrac{256}{9(t+4)}\;\;dt$$ $$\dfrac{t^3}{3}+\dfrac{t^2}{2}+21t+\dfrac {625}{9}\log (t-5)-\dfrac{256}{9}\log (t+4)$$ put $t=e^x$ $$\mathbf {Answer}=\dfrac{e^{3x}}{3}+\dfrac{e^{2x}}{2}+21e^x+\dfrac {625}{9}\log (e^x-5)-\dfrac{256}{9}\log (e^x+4)+C\;$$ solution of partial fraction used above $$\dfrac{41t+420}{t^2 - t - 20}=\dfrac {A}{t-5}+\dfrac{B}{t+4}$$ $${41t+420}={A}(t+4)+B({t-5})$$ put $t=5$ and then $t=-4$ $${41\times 5+420}={A}(5+4)+B({5-5})$$ $$A=\dfrac {625}{9}$$ $${41\times (-4)+420}={A}(-4+4)+B({-4-5})$$ $$B=\dfrac {-256}{9}$$ $$\dfrac{41t+420}{t^2 - t - 20}=\dfrac {625}{9(t-5)}-\dfrac{256}{9(t+4)}$$

Answer 2

Hint: why not try using synthetic division on this expression to get the sum of a polynomial expression and a remainder term? Then you can integrate using the power rule and apply partial fractions decomposition to the remainder.