Question #312ef

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Question #312ef

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Answer 1 · 2016-09-11T17:45:25

$\frac{{(10 e)}^{x}}{ln (10) + 1} + e^{x + 1} + C$ $(10e)^x/(ln(10)+1)+e^(x+1)+C$

Explanation:

Expanding the integral, it becomes:

$= \int e^{x} 10^{x} d x + \int e^{x + 1} d x$ $=inte^x10^xdx+inte^(x+1)dx$

First working with the second integral, let $u = x + 1$ $u=x+1$ so $d u = d x$ $du=dx$ :

$= \int e^{x} 10^{x} + \int e^{u} d u$ $=inte^x10^x+inte^udu$

$= \int e^{x} 10^{x} + e^{u}$ $=inte^x10^x+e^u$

$= \int e^{x} 10^{x} + e^{x + 1}$ $=inte^x10^x+e^(x+1)$

For the remaining integral, rewrite as follows:

$= \int {(10 e)}^{x} d x + e^{x + 1}$ $=int(10e)^xdx+e^(x+1)$

Notice that $10 e$ $10e$ is just a constant. Use the rule: $\int a^{x} d x = \frac{a^{x}}{ln (a)} + C$ $inta^xdx=a^x/ln(a)+C$

$= \frac{{(10 e)}^{x}}{ln (10 e)} + e^{x + 1}$ $=(10e)^x/ln(10e)+e^(x+1)$

Note that $ln (10 e) = ln (10) + ln (e) = ln (10) + 1$ $ln(10e)=ln(10)+ln(e)=ln(10)+1$ :

$= \frac{{(10 e)}^{x}}{ln (10) + 1} + e^{x + 1} + C$ $=(10e)^x/(ln(10)+1)+e^(x+1)+C$

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Question #312ef

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