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

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!

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Hello, and welcome to Math SE! We're in the process of removing the homework tag, so I've removed it from your question. Beyond that, if you can edit your post to be as clearly formatted as possible, then you'll be more likely to get a constructive response. –  Semiclassical Jul 30 '14 at 20:35
Your proposed substitution is fine (the method suggested is the same; they are just manipulating the integrand so that the substitution is easier). But $\int 1/u^3\,du=\int u^{-3}\,du$ would use a power law, not a logarithmic law. –  David Mitra Jul 30 '14 at 20:35

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 $(9-8x^2)^{-3}$

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

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

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.
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