Anti-derivatives

**coach2uf** · April 9th 2008, 04:58 PM

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

**Mathstud28** · April 9th 2008, 05:01 PM

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

**coach2uf** · April 9th 2008, 05:06 PM

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

**Mathstud28** · April 9th 2008, 05:11 PM

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 $\frac{-1}{x}-\frac{1}{x^2}+\frac{1}{x^3}+C$

**coach2uf** · April 9th 2008, 05:14 PM

How do you get there though?

Originally Posted by Mathstud28

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

**Mathstud28** · April 9th 2008, 05:16 PM

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 $g(x)$

**o_O** · April 9th 2008, 05:18 PM

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

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

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

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?

**coach2uf** · April 9th 2008, 05:26 PM

Originally Posted by o_O

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

Wouldn't it be $u^{-4}$ ?

**o_O** · April 9th 2008, 05:34 PM

What do you mean? I just replaced 4t + 9 with u. The exponent -3 remains the same.

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