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.
$\int \frac{f\text{'}\left(x\right)g\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{(g{\left(x\right))}^2}dx=\_\_\_\_$

Short answer

$\int \frac{f\text{'}\left(x\right)g\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{(g{\left(x\right))}^2}dx=\left(\frac{f\left(x\right)}{g\left(x\right)}\right)+C$

Step-by-step solution

Step 1. Given Information

$\int \frac{f\text{'}\left(x\right)g\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{(g{\left(x\right))}^2}dx=\_\_\_\_$

Step 2. Solving the expression

As quotient rule of derivative is$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u\text{'}v-uv\text{'}}{v^2}$

So,

role="math" localid="1649792911630" width="242" height="47">ddxf(x)g(x)=f(x)'g(x)-f(x)g'(x)(g(x))2
role="math" localid="1649792902040" width="258" height="47">df(x)g(x)=f(x)'g(x)-f(x)g'(x)(g(x))2dx

Integrating both sides

$\int d\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\int \left(\frac{f\left(x\right)\text{'}g\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{(g{\left(x\right))}^2}\right)dx$
$\int \frac{f\left(x\right)\text{'}g\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{(g{\left(x\right))}^2}dx=\left(\frac{f\left(x\right)}{g\left(x\right)}\right)+C$