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anonymous

  • one year ago

can someone please explain how the integral of 1/(1+2x)^2 = -1/2(2x+1)

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  1. Empty
    • one year ago
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    Yeah, make the substitution: \(1+2x = u\) then differentiate this to get: \(2 = \frac{du}{dx}\) multiply both sides of this by dx to get: \(2dx=du\) Now divide both sides by 2: \(dx = \frac{1}{2}du\) So when you substitute into your integral: \(\int \frac{1}{(1+2x)^2} dx = \int \frac{1}{u^2} \frac{1}{2}du\) Now it should be a little easier to solve from here, then you can substitute back in u=1+2x when you're done.

  2. anonymous
    • one year ago
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    omg thank you so much i understand it now

  3. Empty
    • one year ago
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    Technically you're not allowed to multiply by "dx" like that, but it won't really be a problem until you get to multivariable calculus.

  4. ganeshie8
    • one year ago
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    After doing few problems, it becomes easy to guess these directly notice that the derivative of \(\dfrac{1}{x}\) is \(-\dfrac{1}{x^2}\), so \(-\dfrac{1}{x^2}\) is an antiderivative of \(\dfrac{1}{x}\)

  5. ganeshie8
    • one year ago
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    similarly, if you see that the derivative of \(\dfrac{1}{x+1}\) is \(-\dfrac{1}{(x+1)^2}\), you can say immediately that \(-\dfrac{1}{(x+1)^2}\) is an antiderivative of \(\dfrac{1}{x+1}\)

  6. anonymous
    • one year ago
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    that is cool stuff to know this will help out alot

  7. ganeshie8
    • one year ago
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    it is called "advanced guessing", becomes easy with some practice

  8. Empty
    • one year ago
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    Lol "advanced guessing"

  9. Empty
    • one year ago
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    Is that really called that or are you just joking around XD

  10. anonymous
    • one year ago
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    thats funny

  11. ganeshie8
    • one year ago
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    Haha here is my theory : guessing \(\cos x\) is an antiderivative of \(\sin x\) is just "plain guessing"

  12. ganeshie8
    • one year ago
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    but guessing \(e^{2x}\) is an antiderivative of \(2e^{2x}\) is "advanced guessing" as we're jumping ahead avoiding u substitution

  13. anonymous
    • one year ago
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    that is awesome by that theory i have integral of 1/(2x+1)^2 = -1/(2x+1) that is sooo awesome

  14. ganeshie8
    • one year ago
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    i do remember reading that phrase "advanced guessing" from some textbook :)

  15. ganeshie8
    • one year ago
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    Yes! I see that you're getting hang of it, but you will need to be somewhat careful here your guess is a good one, the antiderivative of `1/(2x+1)^2` should look something like `-1/(2x+1) `

  16. ganeshie8
    • one year ago
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    but thats only the first step, to be sure, differentiate `-1/(2x+1) ` and see if you really get ` 1/(2x+1)^2`

  17. Empty
    • one year ago
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    Yeah, true integrals can be hard but derivatives are easy. So whenever you solve an integral, you should try to check yourself by just differentiating it. It's a useful trick and you'll get better at derivatives in the process so it's kind of like a win-win situation haha.

  18. anonymous
    • one year ago
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    ok thats right i see that is important i need to keep that in mind that checking the derivative at the end is important

  19. anonymous
    • one year ago
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    thank you guys for all the help, i gotta get to bed

  20. ganeshie8
    • one year ago
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    gnite!

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