Q. 53 - Calculus

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

Step-by-Step Solution

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 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,

width="242" height="47" style="max-width: none; vertical-align: -19px;" $\frac{d}{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{f\left(x\right)\text{'}g\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{(g{\left(x\right))}^2}$
width="258" height="47" style="max-width: none; vertical-align: -19px;" $d\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\left(\frac{f\left(x\right)\text{'}g\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{(g{\left(x\right))}^2}\right)dx$

Integrating both sides 

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