Calculate $\int 5^{x+1}e^{2x-1}\,dx$ How to calculate following integration?
$$\int 5^{x+1}e^{2x-1}dx$$
 A: Use integration by parts. Set $2x=y$ and you'll get:
$$5e^{-1}\int5^{\frac{y}{2}}e^ydy=5e^{-1}\int5^{\frac{y}{2}}d(e^y)$$
After that just a little effort. :)
$$5e^{-1}\int5^{\frac{y}{2}}e^ydy=5e^{-1}\int5^{\frac{y}{2}}d(e^y)=5e^{-1}(5^{\frac{y}{2}}e^y-\frac{\ln(5)}{2}\int5^{\frac{y}{2}}e^ydy)$$
Combining the first and the last parts:
$$5e^{-1}\int5^{\frac{y}{2}}e^ydy=5e^{-1}(5^{\frac{y}{2}}e^y-\frac{\ln(5)}{2}\int5^{\frac{y}{2}}e^ydy)$$
Cancelling out multipliers and collecting the integral terms:
$$(1+\frac{\ln(5)}{2})\int5^{\frac{y}{2}}e^ydy=5^{\frac{y}{2}}e^y$$
So one shoulf get:
$$\int5^{\frac{y}{2}}e^ydy=\frac{5^{\frac{y}{2}}e^y}{1+\frac{\ln(5)}{2}}$$
So after that just remember the multiplier $5e^{-1}$ and the original variable $2x=y$
A: $$\int 5^{x+1}e^{2x-1}dx=\int e^{(x+1)\ln 5}e^{2x-1}dx=\int e^{(2+\ln 5)x+\ln5-1}dx=5e^{-1}\int e^{(2+\ln 5)x}dx$$
Can you continue?
A: $$\displaystyle \int 5^{x+1}e^{2x-1}dx$$
$$=\int e^{(x+1)ln(5)}e^{2x-1}dx$$
$$=\int e^{(x+1)ln(5)+2x-1}dx$$
$$=\int e^{(\ln(5)+2)x+\ln(5)-1}dx$$
$$=\int e^{(\ln(5)+2)x}\cdot e^{\ln(5)-1}dx$$
$$=\frac{5}{e}\int e^{(\ln(5)+2)x}dx$$
$$=\frac{5}{e}\int e^{(\ln(5)+2)x}dx$$
$$=\frac{5^{x} e^{2 x}}{e \log{\left (5 \right )} + 2 e}$$
A: $$\int 5^{x+1}e^{2x-1}dx=\int\frac{5}{e}(5e^2)^xdx$$
$$(5e^2)^x=u\Rightarrow(5e^2)^xdx=\frac{du}{\ln(5e^2)}$$
$$\int\frac{5}{e}(5e^2)^xdx=\frac{5}{e\ln(5e^2)}\int du=\frac{5}{e\ln(5e^2)}(u+C)$$
A: A computerised calculation of your problem
