anonymous
  • anonymous
Need Help! Evaluate the integral (1-x/x)^2 *dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1439271323768:dw|
ganeshie8
  • ganeshie8
whats stopping you from expanding and working each term ?
anonymous
  • anonymous
So I did it, but I got stuck on one part

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anonymous
  • anonymous
Let me draw it
anonymous
  • anonymous
|dw:1439271526217:dw|
anonymous
  • anonymous
What happens after this...because the solution manual is saying something else
ganeshie8
  • ganeshie8
looks good! expand that out using the identity \[(a-b)^2=a^2-2ab+b^2\]
ganeshie8
  • ganeshie8
|dw:1439271609165:dw|
ganeshie8
  • ganeshie8
\[\left(\frac{1}{x}-1\right)^2=?\]
anonymous
  • anonymous
OH thats what they did
anonymous
  • anonymous
okay, I totally forgot about the identity and was wondering how they came up with the middle term
ganeshie8
  • ganeshie8
it helps to review basic algebraic and trig identities while doing integrals
anonymous
  • anonymous
I would honestly love doing integrals....but theres just so much algebra
ganeshie8
  • ganeshie8
Exactly! calculus is always easy, its just the algebra part that is hard
ganeshie8
  • ganeshie8
its not hard as such, its just pain
anonymous
  • anonymous
lol tell me about it, I have my final in a few days..and im freaking out Either way, how is the middle term -2/x , im getting +2/x
ganeshie8
  • ganeshie8
\[(a-b)^2=a^2\color{red}{-}2ab+b^2\]
anonymous
  • anonymous
So it would be -2*(1/x)*(-1), right?
ganeshie8
  • ganeshie8
\[\left(\frac{1}{x}-1\right)^2=\frac{1}{x^2}-\frac{2}{x}+1\]
anonymous
  • anonymous
But the middle part is -2*(1/x)*(-1), so wouldnt that make it +2/x instead of -2/x?
ganeshie8
  • ganeshie8
\[\left(\frac{1}{x}-1\right)^2=(\frac{1}{x})^2-2(\frac{1}{x})(1)+1^2\]
anonymous
  • anonymous
why dont we multiply by negative, because isnt that the b value?
ganeshie8
  • ganeshie8
\(a = \frac{1}{x}\) \(b=1\)
anonymous
  • anonymous
kk, I get it
ganeshie8
  • ganeshie8
:)
ganeshie8
  • ganeshie8
don't forget to integrate the terms
anonymous
  • anonymous
btw are you in Uni?
anonymous
  • anonymous
Also, how do we integrate 1/x^2?
Astrophysics
  • Astrophysics
\[\frac{ 1 }{ x^2 } = x^{-2} ~~~~~\int\limits \frac{ 1 }{ x^2 } dx = \int\limits x^{-2} dx = \frac{ x^{-1} }{ -1 }+C = - \frac{ 1 }{ x }+C\]
anonymous
  • anonymous
nvm, I watched this video https://www.youtube.com/watch?v=ZFZRid582Lk, and even though its in spanish...I figured it out, thanks though astro
ganeshie8
  • ganeshie8
here is a trick question, \[\int \frac{1}{x}~dx = ?\] @mary_am
anonymous
  • anonymous
lnIxI + c
ganeshie8
  • ganeshie8
nvm that video is a spoiler
Astrophysics
  • Astrophysics
lol
anonymous
  • anonymous
k im not dumb.. I knew what ∫1x dx was
anonymous
  • anonymous
*THAT
ganeshie8
  • ganeshie8
Ofcourse you're not, I see :)
Astrophysics
  • Astrophysics
I am dumb, and yes I am xD
Astrophysics
  • Astrophysics
Sounds about right
Astrophysics
  • Astrophysics
Eihsenag university
Astrophysics
  • Astrophysics
We enjoy helping people, it's a good way to learn ourselves :-)
Astrophysics
  • Astrophysics
They intertwine
Astrophysics
  • Astrophysics
Thank you and you to!

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