1. Anti-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

2. ok

Originally Posted by coach2uf
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 $\int{(4t-9)^3}dt$...now we can rewrite this as $\frac{1}{4}\int{(4t-9)^3}dt$....and now using the rules of $\int{f(g(x))\cdot{g'(x)}$ we can see the answer easily

and you also have $\int{x^2+2x-3}{x^4}$...split the fraction apart to get $\frac{x^2}{x^4}+\frac{2x}{x^4}-\frac{3}{x^4}$ and the answer should be readily apparent

3. That's just it; I can't see the answer.

Originally Posted by Mathstud28
you have $\int{(4t-9)^3}dt$...now we can rewrite this as $\frac{1}{4}\int{(4t-9)^3}dt$....and now using the rules of $\int{f(g(x))\cdot{g'(x)}$ we can see the answer easily

and you also have $\int{x^2+2x-3}{x^4}$...split the fraction apart to get $\frac{x^2}{x^4}+\frac{2x}{x^4}-\frac{3}{x^4}$ and the answer should be readily apparent

4. Originally Posted by coach2uf
That's just it; I can't see the answer.
the first answer is $\frac{(4t-9)^4}{16}+C$...the second answer is $\frac{-1}{x}-\frac{1}{x^2}+\frac{1}{x^3}+C$

5. How do you get there though?

Originally Posted by Mathstud28
the first answer is $\frac{(4t-9)^4}{16}+C$...the second answer is $\frac{-1}{x}-\frac{1}{x^2}+\frac{1}{x^3}+C$

6. ok

Originally Posted by coach2uf
How do you get there though?
this is something you will have to learn but I will try to help $\int{ax^n}dx=anx^{n-1}$...and if you have $\int{f(g(x))\cdot{g'(x)}}$ you integrate it as though it was just $f(x)$ and then replace the $x$ with $g(x)$

7. $\int(4t-9)^{-3}dt$

Let $u = 4t - 9$. So: $du = 4 dt \Rightarrow dt = \frac{du}{4}$

$\int(4t - 9)^{-3}dt = \int u^{-3} \cdot \frac{du}{4} = \frac{1}{4} \int u^{-3} du$

And this should be a standard one.

For the second one, what MathStud was saying was to split the fraction up:

$\int \frac{x^{2} - 2x + 3}{x^{4}} dx = \int \left( \frac{x^{2}}{x^{4}} - \frac{2x}{x^{4}} + \frac{3}{x^{4}}\right) dx$

See how each term cancels into something you can integrate rather easily?

8. Originally Posted by o_O
$\int(4t - 9)^{-3}dt = \int u^{-3} \cdot \frac{du}{4} = \frac{1}{4} \int u^{-3} du$
Wouldn't it be $u^{-4}$ ?