Chapter 5.8, Problem 1E

To determine

To Calculate: ${\displaystyle \int \left(\frac{8x^2-4x+5}{x^4}\right)dx}$

Answer to Problem 1E

  ${\displaystyle \int \left(\frac{8x^2-4x+5}{x^4}\right)dx}=-\frac{8}{x}+\frac{2}{x^2}-\frac{5}{3x^3}+C$

Explanation of Solution

Given information:

  ${\displaystyle \int \left(\frac{8x^2-4x+5}{x^4}\right)dx}$

Concept and Formula Used:

  ${\displaystyle \int dx=x+C},{\displaystyle \int \left(x^n\right)dx}=\frac{x^{n+1}}{n+1}+C$ .

Calculation:

It is given that

  $\begin{array}{l}{\displaystyle \int \left(\frac{8x^2-4x+5}{x^4}\right)dx}\\ {\displaystyle \int \left(\frac{8x^2-4x+5}{x^4}\right)dx}={\displaystyle \int \left(\frac{8x^2}{x^4}\right)dx}-{\displaystyle \int \left(\frac{4x}{x^4}\right)dx}+{\displaystyle \int \left(\frac{5}{x^4}\right)dx}\\ {\displaystyle \int \left(\frac{8x^2-4x+5}{x^4}\right)dx}=8{\displaystyle \int \left(x^{-2}\right)dx}-4{\displaystyle \int \left(x^{-3}\right)dx}+5{\displaystyle \int \left(x^{-4}\right)dx}\\ \text{Using }{\displaystyle \int \left(x^n\right)dx}=\frac{x^{n+1}}{n+1}+C\\ {\displaystyle \int \left(\frac{8x^2-4x+5}{x^4}\right)dx}=8\times \frac{x^{-2+1}}{-2+1}-4\times \frac{x^{-3+1}}{-3+1}+5\frac{x^{-4+1}}{-4+1}+C\\ {\displaystyle \int \left(\frac{8x^2-4x+5}{x^4}\right)dx}=-\frac{8}{x}+\frac{2}{x^2}-\frac{5}{3x^3}+C\end{array}$

Conclusion:

  ${\displaystyle \int \left(\frac{8x^2-4x+5}{x^4}\right)dx}=-\frac{8}{x}+\frac{2}{x^2}-\frac{5}{3x^3}+C$