# 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$Done7^(2*x+3)/(2*ln(7))+C