anonymous
  • anonymous
1) x^2+1/6x-9x^2 dx 2) xdx/2+6x+9x^2 3) 1/x^2-1 dx
Calculus1
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
What are you supposed to to?
anonymous
  • anonymous
Find antiderivative
anonymous
  • anonymous
Oh

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anonymous
  • anonymous
An antiderivative of a function f is a function whose derivative is f . In other words, F is an antiderivative of f if F' = f . To find an antiderivative for a function f , we can often reverse the process of differentiation. For example, if f = x^4 , then an antiderivative of f is F = 1/5 x^5 , which can be found by reversing the power rule. Notice that not only is 1/5 x^5 an antiderivative of f , but so are 1/5 x^5 + 4 , 1/5 x^5 + 6 , etc. In fact, adding or subtracting any constant would be acceptable.
anonymous
  • anonymous
That I get, but I just don't know how we antideriverate those functions
anonymous
  • anonymous
Ok, First try to antideriverate them without the numbers.
anonymous
  • anonymous
Also to do it part by part. So the first one do the numirator and then the denominater.
anonymous
  • anonymous
Just do the same thing on all three.
anonymous
  • anonymous
What, I don' get what u mean?
anonymous
  • anonymous
For the first one do x^2+1 After that do 6x-9x^2
anonymous
  • anonymous
For the second one do x Then do 2+6x+9x^2
anonymous
  • anonymous
Then for the last one do 1first Then do x^2-1
anonymous
  • anonymous
So... The first one will be \[\frac{ \frac{ x^3 }{ 3 }+x }{ 3x^2 -3x^3}\] The second one should be \[\frac{ \frac{x^2 }{ 2 } }{ 2x+3x^2 +3x^3}\] And the third one should be \[\frac{ x }{ \frac{ x^3 }{ 3 } -1 }\]
IrishBoy123
  • IrishBoy123
first one is \( \int \dfrac{x^2+1}{6x-9x^2} dx\) right?? if so you need to use some algebra, eg partial fractions, to simplify that and make it fit some known integration patterns i'd just pile in with with long division as this stuff can get really, really tedious after a while so you have \(\dfrac{x^2+1}{6x-9x^2} = \dfrac{1}{x}\dfrac{x^2+1}{6-9x}\) and why not try \((x^2 + 1) \div (-9x + 6)\) in ye olde long division format
anonymous
  • anonymous
Oh lawd jesus. ain't nobody got time for that!
anonymous
  • anonymous
LOL
anonymous
  • anonymous
the first on i say
IrishBoy123
  • IrishBoy123
PS this is a really good resource http://www.mathsisfun.com/algebra/partial-fractions.html

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