How do you find the antiderivative of #(x^5-x^3+2x)/x^4#?

Question

Paisley Johnson · Accepted Answer

#1/2x^2-lnx-x^-2+c# #"divide the rational function term by term"# #rArrx^5/x^4-x^3/x^4+(2x)/x^4=x-x^-1+2x^-3# #"integrate each term using the "color(blue)"power rule"# #int(ax)^n=a/(n+1)x^(n+1)to(n!=-1)# #"note that "int(x^-1)=int(1/x)=lnx# #rArrint(x-x^-1+2x^-3)dx# #=1/2x^2-lnx-x^-2+c# #"where c is the constant of integration"#

Samantha Adkison · Answer

undefined To find the antiderivative of $(x^5 - x^3 + 2x)/x^4$, you can split the expression into three separate terms: $x^5/x^4$, $-x^3/x^4$, and $2x/x^4$. Then, apply the power rule for integration, which states that $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, where $n$ is any real number except -1, and $C$ is the constant of integration. After integrating each term separately, you can combine the results.

$$
\begin{align*}
\int \frac{x^5 - x^3 + 2x}{x^4} \, dx &= \int \left( \frac{x^5}{x^4} - \frac{x^3}{x^4} + \frac{2x}{x^4} \right) \, dx \
&= \int (x^{5-4} - x^{3-4} + 2x^{-3}) \, dx \
&= \int (x - x^{-1} + 2x^{-3}) \, dx \
&= \frac{x^2}{2} - \ln|x| - \frac{2}{x^2} + C,
\end{align*}
$$
where $C$ is the constant of integration.