Simple anti-derivative problems. I missed the lecture on this and need to be walked through this:
Indefinite integral for: (4t-9)^-3
and
(x^2+2x-3)/(x^4)
thanks a lot
you have $\displaystyle \int{(4t-9)^3}dt$...now we can rewrite this as $\displaystyle \frac{1}{4}\int{(4t-9)^3}dt$....and now using the rules of $\displaystyle \int{f(g(x))\cdot{g'(x)}$ we can see the answer easily
and you also have $\displaystyle \int{x^2+2x-3}{x^4}$...split the fraction apart to get $\displaystyle \frac{x^2}{x^4}+\frac{2x}{x^4}-\frac{3}{x^4}$ and the answer should be readily apparent
this is something you will have to learn but I will try to help $\displaystyle \int{ax^n}dx=anx^{n-1}$...and if you have $\displaystyle \int{f(g(x))\cdot{g'(x)}}$ you integrate it as though it was just $\displaystyle f(x)$ and then replace the $\displaystyle x$ with $\displaystyle g(x)$
$\displaystyle \int(4t-9)^{-3}dt$
Let $\displaystyle u = 4t - 9$. So: $\displaystyle du = 4 dt \Rightarrow dt = \frac{du}{4}$
So performing your substitution:
$\displaystyle \int(4t - 9)^{-3}dt = \int u^{-3} \cdot \frac{du}{4} = \frac{1}{4} \int u^{-3} du$
And this should be a standard one.
For the second one, what MathStud was saying was to split the fraction up:
$\displaystyle \int \frac{x^{2} - 2x + 3}{x^{4}} dx = \int \left( \frac{x^{2}}{x^{4}} - \frac{2x}{x^{4}} + \frac{3}{x^{4}}\right) dx$
See how each term cancels into something you can integrate rather easily?