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What is $\int \frac{7}{8}﻿{e}^{2t-1}dt$?
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drenkttj9
Let u=2t-1. Then du=2dt, so $\frac{1}{2}du=dt$. Rewrite using u and du.
$\int \frac{7}{8}{e}^{u}\frac{1}{2}du$
Simplify.
$\int \frac{7{e}^{u}}{16}du$
Since $\frac{7}{16}$ is constant with respect to u, move $\frac{7}{16}$ out of the integral.
$\frac{7}{16}\int {e}^{u}du$
The integral of ${e}^{u}$ with respect to u is ${e}^{u}$
$\frac{7}{16}\left({e}^{u}+C\right)$
Simplify.
$\frac{7}{16}{e}^{u}+C$
Replace all occurrences of u with 2t-1.
$\frac{7}{16}{e}^{2t-1}+C$
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