Problem of the Week

Updated at Nov 25, 2013 5:02 PM

How can we solve for the integral of 72x+3{7}^{2x+3}?

Below is the solution.



∫72x+3 dx\int {7}^{2x+3} \, dx

1
Use Integration by Substitution.
Let u=2x+3u=2x+3, du=2 dxdu=2 \, dx, then dx=12 dudx=\frac{1}{2} \, du

2
Using uu and dudu above, rewrite ∫72x+3 dx\int {7}^{2x+3} \, dx.
∫7u2 du\int \frac{{7}^{u}}{2} \, du

3
Use Constant Factor Rule: ∫cf(x) dx=c∫f(x) dx\int cf(x) \, dx=c\int f(x) \, dx.
12∫7u du\frac{1}{2}\int {7}^{u} \, du

4
Use this property: ∫ax dx=axln⁡a\int {a}^{x} \, dx=\frac{{a}^{x}}{\ln{a}}.
7u2ln⁡7\frac{{7}^{u}}{2\ln{7}}

5
Substitute u=2x+3u=2x+3 back into the original integral.
72x+32ln⁡7\frac{{7}^{2x+3}}{2\ln{7}}

6
Add constant.
72x+32ln⁡7+C\frac{{7}^{2x+3}}{2\ln{7}}+C

Done
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