anonymous
  • anonymous
can someone please explain how the integral of 1/(1+2x)^2 = -1/2(2x+1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Empty
  • Empty
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.
anonymous
  • anonymous
omg thank you so much i understand it now
Empty
  • Empty
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.

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ganeshie8
  • ganeshie8
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 \(\dfrac{1}{x}\)
ganeshie8
  • ganeshie8
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 \(\dfrac{1}{x+1}\)
anonymous
  • anonymous
that is cool stuff to know this will help out alot
ganeshie8
  • ganeshie8
it is called "advanced guessing", becomes easy with some practice
Empty
  • Empty
Lol "advanced guessing"
Empty
  • Empty
Is that really called that or are you just joking around XD
anonymous
  • anonymous
thats funny
ganeshie8
  • ganeshie8
Haha here is my theory : guessing \(\cos x\) is an antiderivative of \(\sin x\) is just "plain guessing"
ganeshie8
  • ganeshie8
but guessing \(e^{2x}\) is an antiderivative of \(2e^{2x}\) is "advanced guessing" as we're jumping ahead avoiding u substitution
anonymous
  • anonymous
that is awesome by that theory i have integral of 1/(2x+1)^2 = -1/(2x+1) that is sooo awesome
ganeshie8
  • ganeshie8
i do remember reading that phrase "advanced guessing" from some textbook :)
ganeshie8
  • ganeshie8
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) `
ganeshie8
  • ganeshie8
but thats only the first step, to be sure, differentiate `-1/(2x+1) ` and see if you really get ` 1/(2x+1)^2`
Empty
  • Empty
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.
anonymous
  • anonymous
ok thats right i see that is important i need to keep that in mind that checking the derivative at the end is important
anonymous
  • anonymous
thank you guys for all the help, i gotta get to bed
ganeshie8
  • ganeshie8
gnite!

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