How would I go about evaluating $\int \frac{x}{(9-8x^2)^3}dx$?

Question

So I have homework on webAssign (a site used by my college), and I am not understanding the logic as to why I am taking the steps into solving the integral it is telling me to take. So I'll list the website's steps and if someone could tell how/why it works step by step, I'd really appreciate it. $$\int \left( \frac{x}{(9-8x^2)^3} \right) dx$$ So it tell's me the the derivative of $9-8x^2=-16x$ and to let $g(x)=9-8x^2$. $$$$Then it proceeds to tell me to multiply and divide the integrand by $-16$ and rewrite the integral like-so $$\int \left( \frac{x}{(9-8x^2)^3} \right) dx=-\frac{1}{16}\int(9-8x^2)^{-3}(-16x)dx$$ That step is where I get completely lost and I don't now how I went from the original integral to the new integral. I cannot go further since I do not know the next step. What I thoughI could in this problem is use $u=9-8x^2$ and $du=-16xdx$ which then $\frac{-1}{16}du=xdx$ Which would give me $\frac{-1}{16}\int\frac{du}{u^3}$ and then use logarithmic rules of integrals to solve, but I suppose it isn't right, so how would I go about correctly solving that problem? Thanks for all the help in advance NOTE: The steps take on the first method are not taken by my choice, so I do not know why they are done, so a step by step explanation on how to solve would be greatly appreciated. Thanks!

Answer 1

What happened is the numerator and denominator got multiplied by $-16$ to obtain

$$\int\dfrac{-16xdx}{-16(9-8x^2)^3}$$

You may recognize the numerator now as $du$. So this can be simply rewrittten as

$$\int\dfrac{du}{-16u^3}$$

Instead however, they pulled the constant $-\frac1{16}$ outside the integral and rewrote $\dfrac1{(9-8x^2)^3}$ as $(9-8x^2)^{-3}$

Answer 2

$\int u^{-3} du$ doesnt need logarithms to be solved.

$\int x^n dx = \frac{x^{n+1}}{n+1} + C$

https://www.youtube.com/watch?v=qclrs-1rpKI

Answer 3

I am not the best at this but it appears to be a shortcut. You have multiplied by a constant that is equivalent to 1: $$\frac{-16}{-16}=1$$ and then taken out $\frac{-1}{16}$. This is helpful as setting $u=9-8x^2$, you have: $$\frac{du}{dx}=-16x$$ Which should make integration by substitution easier

Answer 4

This can be solved via substitution. Here are the steps \[ \int \frac{x}{(9-8x^2)^3} dx \] Let $u=9-8x^2$, then \[ \frac{du}{dx}=\frac{d}{dx}[9-8x^2]=0-16x=-16x \] This also implies that \[ du=-16x\ dx\Rightarrow -\frac{1}{16}\ du=x\ dx \] So now we have \[ \int \frac{x}{(9-8x^2)^3} dx = -\frac{1}{16}\int \frac{1}{u^3}\ du = -\frac{1}{16}\int u^{-3}\ du= -\frac{1}{16}\cdot \left(-\frac{1}{2}u^{-2}\right)+C=\frac{1}{32}u^{-2}+C \] Thus \[ \frac{1}{32}u^{-2}+C= \frac{1}{32}(9-8x^2)^{-2}+C \] I hope this helps you understand.