Find the indefinite integral of 2^x times (2^x + 4)^5 with respect to x.

Step-by-step solution

1. 1

   Look for an inside function whose derivative is already sitting outside. The bracket $2^x+4$ is raised to a power, and its derivative is

   $\frac{d}{dx}\left(2^{x}+4\right)=2^{x}\ln 2,$

   which is the outside factor $2^x$ up to the constant $\ln 2$. That is the signature of a substitution problem.

2. 2

   Substitute. Let

   $u=2^{x}+4\quad\Longrightarrow\quad du=2^{x}\ln 2\,dx\quad\Longrightarrow\quad 2^{x}\,dx=\frac{du}{\ln 2}.$

   The base here is $2$, not $e$, so the $\ln 2$ must be carried — dropping it is the single most common error in this integral.

3. 3

   Rewrite the integral entirely in u.

   $\int 2^{x}\left(2^{x}+4\right)^{5}dx=\int u^{5}\,\frac{du}{\ln 2}=\frac{1}{\ln 2}\int u^{5}\,du.$

4. 4

   Apply the power rule and substitute back.

   $\frac{1}{\ln 2}\cdot\frac{u^{6}}{6}+C=\frac{\left(2^{x}+4\right)^{6}}{6\ln 2}+C.$

5. 5

   Differentiate to confirm. By the chain rule,

   $\frac{d}{dx}\frac{\left(2^{x}+4\right)^{6}}{6\ln 2}=\frac{6\left(2^{x}+4\right)^{5}\cdot 2^{x}\ln 2}{6\ln 2}=2^{x}\left(2^{x}+4\right)^{5},$

   which is the original integrand, so the antiderivative is correct.