Question

Integrate

(e^3x + 2e^2x)/(e^3x + 3e^2x + 9) dx

from 0 to 1.

Answer 1

\[\int_0^1\frac{e^{3x}+2e^{2x}}{e^{3x}+3e^{2x}+9}~dx\]
Let \(u=e^{3x}+3e^{2x}+9\), then \(du=\left(3e^{3x}+6e^{2x}\right)~dx\)
\[\frac{1}{3}\int_0^1\frac{3e^{3x}+6e^{2x}}{e^{3x}+3e^{2x}+9}~dx={\large\frac{1}{3}\int_{13}^{e^3+3e^2+9}\frac{du}{u}}\]