Chapter 4.6, Problem 65E

Occasionally, integration by parts yields an integral of the form ${\displaystyle \int udv}$ that is identical to the original. In some cases, we can then solve for ${\displaystyle \int udv}$ algebraically. For example, to find ${\displaystyle \int 2^x{e}^x\text{ we Let }u=2^x\text{ and }dv=e^x\text{ so }du=\left(\text{In2}\right)2^{x}dx}\text{ and }v=e^x$. Using integral by parts, we have

${\displaystyle \int 2^x{e}^{x}dx=2^x{e}^x-\text{In }2{\displaystyle \int 2^x{e}^{x}dx}}$.

Note that ${\displaystyle \int 2^x{e}^{x}dx}$ appears twice. Adding $\text{In }2{\displaystyle \int 2^x{e}^{x}dx}$ to both sides, we have

${\displaystyle \int 2^x{e}^{x}dx+\text{In 2}}{\displaystyle \int 2^x{e}^{x}dx=2^x{e}^x}$

$\left(1+\text{In 2}\right){\displaystyle \int 2^x{e}^{x}dx}=2^x{e}^x$

${\displaystyle \int 2^x{e}^{x}dx=\frac{2^x{e}^x}{1+\text{In 2}}+C}$.

Use this method to evaluate the integrals in Exercises 59-62.

${\displaystyle \int 3^x{e}^{x}dx}$