Problem of the Week

Updated at Nov 25, 2013 5:02 PM

How can we solve for the integral of \({7}^{2x+3}\)?

Below is the solution.



\[\int {7}^{2x+3} \, dx\]

1
Use Integration by Substitution.
Let \(u=2x+3\), \(du=2 \, dx\), then \(dx=\frac{1}{2} \, du\)

2
Using \(u\) and \(du\) above, rewrite \(\int {7}^{2x+3} \, dx\).
\[\int \frac{{7}^{u}}{2} \, du\]

3
Use Constant Factor Rule: \(\int cf(x) \, dx=c\int f(x) \, dx\).
\[\frac{1}{2}\int {7}^{u} \, du\]

4
Use this property: \(\int {a}^{x} \, dx=\frac{{a}^{x}}{\ln{a}}\).
\[\frac{{7}^{u}}{2\ln{7}}\]

5
Substitute \(u=2x+3\) back into the original integral.
\[\frac{{7}^{2x+3}}{2\ln{7}}\]

6
Add constant.
\[\frac{{7}^{2x+3}}{2\ln{7}}+C\]

Done
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