Consider the integral $\int \frac{x^{-2}-4}{x^3}dx$.

(a) Solve this integral by using u-substitution.

(b) Solve the integral another way, using algebra to simplify the integrand first.

(c) How must your two answers be related? Use algebra to prove this relationship.

Consider the integral $\int \frac{x^{-2}-4}{x^3}dx$.

(a) Solve this integral by using u-substitution.

(b) Solve the integral another way, using algebra to simplify the integrand first.

(c) How must your two answers be related? Use algebra to prove this relationship.

Let

$$\begin{gathered} u=x^{-2}-4 \\ \frac{du}{dx}=-2x^{-3} \\ \frac{du}{dx}=-2\frac{1}{x^3} \\ -\frac{1}{2}du=\frac{1}{x^3}dx \end{gathered}$$

$$\begin{gathered} \int \frac{x^{-2}-4}{x^3}dx=-\frac{1}{2}\int udu \\ \int \frac{x^{-2}-4}{x^3}dx=-\frac{1}{2}\left[\frac{u^{1+1}}{1+1}\right]+C \\ \int \frac{x^{-2}-4}{x^3}dx=-\frac{1}{2}\left[\frac{u^2}{2}\right]+C \\ \int \frac{x^{-2}-4}{x^3}dx=-\frac{u^2}{4}+C \\ \int \frac{x^{-2}-4}{x^3}dx=-\frac{{(x^{-2}-4)}^2}{4}+C \end{gathered}$$

$$\begin{gathered} \int \frac{x^{-2}-4}{x^3}dx=\int \frac{1}{x^3}\left(x^{-2}-4\right)dx \\ \int \frac{x^{-2}-4}{x^3}dx=\int \frac{1}{x^3}\left(\frac{1}{x^2}-4\right)dx \\ \int \frac{x^{-2}-4}{x^3}dx=\int \left(\frac{1}{x^3}·\frac{1}{x^2}-\frac{1}{x^3}·4\right)dx \\ \int \frac{x^{-2}-4}{x^3}dx=\int \left(\frac{1}{x^5}-\frac{4}{x^3}\right)dx \end{gathered}$$

$$\begin{gathered} \int \frac{x^{-2}-4}{x^3}dx=\int \frac{1}{x^5}dx-\int \frac{4}{x^3}dx \\ \int \frac{x^{-2}-4}{x^3}dx=\int x^{-5}dx-4\int x^{-3}dx \\ \int \frac{x^{-2}-4}{x^3}dx=\left[\frac{x^{-5+1}}{-5+1}\right]-4\left[\frac{x^{-3+1}}{-3+1}\right]+C \\ \int \frac{x^{-2}-4}{x^3}dx=\left[\frac{x^{-4}}{-4}\right]-4\left[\frac{x^{-2}}{-2}\right]+C \\ \int \frac{x^{-2}-4}{x^3}dx=-\frac{1}{4x^4}+\frac{2}{x^2}+C \end{gathered}$$

The value of integral in part (a) is

$\int \frac{x^{-2}-4}{x^3}dx=-\frac{{(x^{-2}-4)}^2}{4}+C$

The value of integral in part (b) is

$\int \frac{x^{-2}-4}{x^3}dx=-\frac{1}{4x^4}+\frac{1}{2x^2}+C$

The value of both answer differ by a constant

$$\begin{gathered} -\frac{{(x^{-2}-4)}^2}{4}=-\left(\frac{{\left(x^{-2}\right)}^2-2\times x^{-2}\times 4+{\left(4\right)}^2}{4}\right) \\ -\frac{{(x^{-2}-4)}^2}{4}=\frac{-x^{-4}+8x^{-2}-16}{4} \\ -\frac{{(x^{-2}-4)}^2}{4}=\frac{-x^{-4}}{4}+\frac{8x^{-2}}{4}-\frac{16}{4} \\ -\frac{{(x^{-2}-4)}^2}{4}=\frac{-x^{-4}}{4}+2x^{-2}-4 \\ -\frac{{(x^{-2}-4)}^2}{4}=-\frac{1}{4x^4}+\frac{2}{x^2}-4 \end{gathered}$$