# Calculate int 5^(x+1)e^(2x-1)dx

Calculate $\int {5}^{x+1}{e}^{2x-1}\phantom{\rule{thinmathspace}{0ex}}dx$
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getrdone07tl
Use integration by parts. Set $2x=y$ and you'll get:
$5{e}^{-1}\int {5}^{\frac{y}{2}}{e}^{y}dy=5{e}^{-1}\int {5}^{\frac{y}{2}}d\left({e}^{y}\right)$
Then
$5{e}^{-1}\int {5}^{\frac{y}{2}}{e}^{y}dy=5{e}^{-1}\int {5}^{\frac{y}{2}}d\left({e}^{y}\right)=5{e}^{-1}\left({5}^{\frac{y}{2}}{e}^{y}-\frac{\mathrm{ln}\left(5\right)}{2}\int {5}^{\frac{y}{2}}{e}^{y}dy\right)$
Combining the first and the last parts:
$5{e}^{-1}\int {5}^{\frac{y}{2}}{e}^{y}dy=5{e}^{-1}\left({5}^{\frac{y}{2}}{e}^{y}-\frac{\mathrm{ln}\left(5\right)}{2}\int {5}^{\frac{y}{2}}{e}^{y}dy\right)$
Cancelling out multipliers and collecting the integral terms:
$\left(1+\frac{\mathrm{ln}\left(5\right)}{2}\right)\int {5}^{\frac{y}{2}}{e}^{y}dy={5}^{\frac{y}{2}}{e}^{y}$
So one shoulf get:
$\int {5}^{\frac{y}{2}}{e}^{y}dy=\frac{{5}^{\frac{y}{2}}{e}^{y}}{1+\frac{\mathrm{ln}\left(5\right)}{2}}$
So after that just remember the multiplier $5{e}^{-1}$ and the original variable $2x=y$
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Kasey Reese
Another way to solve:
$\int {5}^{x+1}{e}^{2x-1}dx=\int \frac{5}{e}\left(5{e}^{2}{\right)}^{x}dx$
$\left(5{e}^{2}{\right)}^{x}=u⇒\left(5{e}^{2}{\right)}^{x}dx=\frac{du}{\mathrm{ln}\left(5{e}^{2}\right)}$
$\int \frac{5}{e}\left(5{e}^{2}{\right)}^{x}dx=\frac{5}{e\mathrm{ln}\left(5{e}^{2}\right)}\int du=\frac{5}{e\mathrm{ln}\left(5{e}^{2}\right)}\left(u+C\right)$