Question

Indefinite integrals of combinations: Fill in the blanks to complete the integration rules that follow. You may assume that $f$and $g$are continuous functions and that$k$ is any real number.
$\int f\text{'}\left(g\right(x\left)\right)g\text{'}\left(x\right)dx=\_\_\_\_\_$

Short answer

$\int f\text{'}\left(g\right(x\left)\right)g\text{'}\left(x\right)dx=f\text{'}\left(g\right(x\left)\right)+C$

Step-by-step solution

Step 1. Given Information

$\int f\text{'}\left(g\right(x\left)\right)g\text{'}\left(x\right)dx=\_\_\_\_\_$

Step 2. Solving the integral

Let $g\left(x\right)=t$
Differentiate with respect to '$t$'
$\frac{d}{dt}g\left(x\right)=\frac{dt}{dt}$

$g\text{'}\left(x\right)dx=dt$
Putting the value$\int f\text{'}\left(g\right(x\left)\right)g\text{'}\left(x\right)dx=\int f\text{'}\left(t\right)dt$

Step 3. Using the antiderivative

As $\int f\left(x\right)dx=F\left(x\right)+C$where $F$is the antiderivative of $f$

Here antiderivative of $f\text{'}\left(x\right)=f\left(x\right)$
$\int f\text{'}\left(g\right(x\left)\right)g\text{'}\left(x\right)dx=f\left(t\right)+C$
Putting $t=g\left(x\right)$
$\int f\text{'}\left(g\right(x\left)\right)g\text{'}\left(x\right)dx=f\left(g\right(x\left)\right)+C$