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

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- Apr 9th 2008, 04:58 PMcoach2ufAnti-derivatives
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 - Apr 9th 2008, 05:01 PMMathstud28ok
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 - Apr 9th 2008, 05:06 PMcoach2uf
- Apr 9th 2008, 05:11 PMMathstud28
- Apr 9th 2008, 05:14 PMcoach2uf
- Apr 9th 2008, 05:16 PMMathstud28ok
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 $\displaystyle g(x)$

- Apr 9th 2008, 05:18 PMo_O
$\displaystyle \int(4t-9)^{-3}dt$

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? - Apr 9th 2008, 05:26 PMcoach2uf
- Apr 9th 2008, 05:34 PMo_O
What do you mean? I just replaced 4t + 9 with u. The exponent -3 remains the same.