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Ian Mateus
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How would I go about solving this integralevaluating $\int \frac{x}{(9-8x^2)^3}dx$?

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. $$\int(\frac{x}{(9-8x^2)^3})dx$$$$\int \left( \frac{x}{(9-8x^2)^3} \right) dx$$ 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(\frac{x}{(9-8x^2)^3})dx=-\frac{1}{16}\int(9-8x^2)^{-3}(-16x)dx$$$$\int \left( \frac{x}{(9-8x^2)^3} \right) dx=-\frac{1}{16}\int(9-8x^2)^{-3}(-16x)dx$$ 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!

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. $$\int(\frac{x}{(9-8x^2)^3})dx$$ 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(\frac{x}{(9-8x^2)^3})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!

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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