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How to calculate following integration?

$$\int 5^{x+1}e^{2x-1}dx$$

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closed as off-topic by user21820, Adrian Keister, YuiTo Cheng, Xander Henderson, José Carlos Santos Jun 7 at 18:29

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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 $2x=y$

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$$\int 5^{x+1}e^{2x-1}dx=\int e^{(x+1)\ln 5}e^{2x-1}dx=\int e^{(2+\ln 5)x+\ln5-1}dx=5e^{-1}\int e^{(2+\ln 5)x}dx$$ Can you continue?

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  • 1
    $\begingroup$ Why the downvote? Do you think that's wrong? $\endgroup$ – Dennis Gulko Mar 25 '13 at 11:01
  • $\begingroup$ I did not down vote your post, but I imagine that some users may consider that by answering this type of questions (with no context or the demonstration of any effort to solve the question in the OP) stimulates a king of question that does not comply with the community standards. $\endgroup$ – PierreCarre Jun 5 at 12:50
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$$\displaystyle \int 5^{x+1}e^{2x-1}dx$$ $$=\int e^{(x+1)ln(5)}e^{2x-1}dx$$ $$=\int e^{(x+1)ln(5)+2x-1}dx$$ $$=\int e^{(\ln(5)+2)x+\ln(5)-1}dx$$ $$=\int e^{(\ln(5)+2)x}\cdot e^{\ln(5)-1}dx$$ $$=\frac{5}{e}\int e^{(\ln(5)+2)x}dx$$ $$=\frac{5}{e}\int e^{(\ln(5)+2)x}dx$$ $$=\frac{5^{x} e^{2 x}}{e \log{\left (5 \right )} + 2 e}$$

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$$\int 5^{x+1}e^{2x-1}dx=\int\frac{5}{e}(5e^2)^xdx$$

$$(5e^2)^x=u\Rightarrow(5e^2)^xdx=\frac{du}{\ln(5e^2)}$$

$$\int\frac{5}{e}(5e^2)^xdx=\frac{5}{e\ln(5e^2)}\int du=\frac{5}{e\ln(5e^2)}(u+C)$$

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  • $\begingroup$ Nice solution Adi $\endgroup$ – mrs Mar 25 '13 at 19:15
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A computerised calculation of your problem

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