If \(\int \frac{2e^x + 3e^{-x}}{4e^x + 7e^{-x}}dx\) = \(\frac{1}{14}\)(ux + v loge(4ex + 7e-x) + C, where C is a constant of integration, then u + v is equal to __________.
\(\int \frac{2e^x}{4e^x + 7e^{-x}}dx\) + 3\(\int \frac{e^x}{4e^x + 7e^{-x}}dx\)
= \(\int \frac{2e^{2x}}{4e^{2x} + 7}dx\) + 3\(\int \frac{e^{-2x}}{4 + 7e^{-2x}}dx\)dx
Let 4e2x + 7 = T
8e2x dx = dT
2e2x dx = \(\frac{dT}{4}\)
Let 4 + 7e-2x = t
-14e-2x dx = dt
e-2x dx = -\(\frac{dt}{14}\)
\(\int \frac{dT}{4T} - \frac{3}{14} \int \frac{dt}{t}\)
= \(\frac{1}{4}\)log T - \(\frac{3}{14}\)log t + C
= \(\frac{1}{4}\) log (4e2x + 7) - \(\frac{3}{14}\)log (4 + 7e-2x) + C
= \(\frac{1}{14}\) [ \(\frac{1}{2}\)log(4ex+ 7e-x) + \(\frac{13}{2}x\) ] + C
u = \(\frac{13}{2}\), v = \(\frac{1}{2}\) ⇒ u + v = 7