Find the Antiderivative f(x)=(1.25^(x/4))^16

$f (x) = {({1.25}^{\frac{x}{4}})}^{16}$

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Set up the integral to solve.

$F (x) = \int {({1.25}^{\frac{x}{4}})}^{16} d x$

Multiply the exponents in ${({1.25}^{\frac{x}{4}})}^{16}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\int {1.25}^{\frac{x}{4} \cdot 16} d x$

Cancel the common factor of $4$ .

Factor $4$ out of $16$ .

$\int {1.25}^{\frac{x}{4} \cdot (4 (4))} d x$

Cancel the common factor.

$\int {1.25}^{\frac{x}{4} \cdot (4 \cdot 4)} d x$

Rewrite the expression.

$\int {1.25}^{x \cdot 4} d x$

Move $4$ to the left of $x$ .

$\int {1.25}^{4 x} d x$

Let $u = 4 x$ . Then $d u = 4 d x$ , so $\frac{1}{4} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

Let $u = 4 x$ . Find $\frac{d u}{d x}$ .

Differentiate $4 x$ .

$\frac{d}{d x} [4 x]$

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$4 \frac{d}{d x} [x]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$4 \cdot 1$

Multiply $4$ by $1$ .

$4$

Rewrite the problem using $u$ and $d u$ .

$\int {1.25}^{u} \cdot \frac{1}{4} d u$

Combine ${1.25}^{u}$ and $\frac{1}{4}$ .

$\int \frac{{1.25}^{u}}{4} d u$

Since $\frac{1}{4}$ is constant with respect to $u$ , move $\frac{1}{4}$ out of the integral.

$\frac{1}{4} \int {1.25}^{u} d u$

The integral of ${1.25}^{u}$ with respect to $u$ is $\frac{{1.25}^{u}}{\ln (1.25)}$ .

$\frac{1}{4} (\frac{{1.25}^{u}}{\ln (1.25)} + C)$

Simplify.

Rewrite $\frac{1}{4} (\frac{{1.25}^{u}}{\ln (1.25)} + C)$ as $\frac{1}{4} \cdot \frac{1}{\ln (1.25)} \cdot {1.25}^{u} + C$ .

$\frac{1}{4} \cdot \frac{1}{\ln (1.25)} \cdot {1.25}^{u} + C$

Multiply $\frac{1}{4}$ by $\frac{1}{\ln (1.25)}$ .

$\frac{1}{4 \ln (1.25)} \cdot {1.25}^{u} + C$

Replace all occurrences of $u$ with $4 x$ .

$\frac{1}{4 \ln (1.25)} \cdot {1.25}^{4 x} + C$

The answer is the antiderivative of the function $f (x) = {({1.25}^{\frac{x}{4}})}^{16}$ .

$F (x) =$ $\frac{1}{4 \ln (1.25)} \cdot {1.25}^{4 x} + C$