Given the integral function below
[tex]\int\frac{-6x}{(4x^2-9)^4}dx[/tex]a) The inside function is
[tex]\begin{gathered} 4x^2-9 \\ \end{gathered}[/tex][tex]\therefore u=4x^2-9[/tex]The derivative of the inside function is
[tex]\begin{gathered} u^{\prime}=8x \\ \frac{du}{dx}=8x \\ dx=\frac{du}{8x} \end{gathered}[/tex]b) Therefore, the integral can be written as:
[tex]\int\frac{-6x}{(4x^2-9)^4}dx\rightarrow\int\frac{-6x}{u^4}\times\frac{du}{8x}[/tex][tex]\rightarrow\int\frac{-3}{4u^4}du\rightarrow-\frac{3}{4}\int u^{-4}du[/tex][tex]\begin{gathered} \rightarrow-\frac{3}{4}(\frac{u^{-4+1}}{-4+1})+C \\ \rightarrow-\frac{3}{4}\times\frac{u^{-3}}{-3}+C \\ \rightarrow\frac{1}{4}\times\frac{1}{u^3}+C \end{gathered}[/tex][tex]\begin{gathered} \frac{1}{4u^3}+C \\ Recall\text{ u=}4x^2-9 \\ \therefore \\ \int\frac{-6x}{(4x^2-9)^4}dx=\frac{1}{4(4x^2-9)^3}+C \end{gathered}[/tex]Final answer:
a) The inside function is 4x² - 9, and the derivative is 8x
b) The integration by substitution is
[tex]\begin{equation*} \frac{1}{4(4x^2-9)^3}+C \end{equation*}[/tex]