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anonymous
 one year ago
can someone please explain how the integral of 1/(1+2x)^2 = 1/2(2x+1)
anonymous
 one year ago
can someone please explain how the integral of 1/(1+2x)^2 = 1/2(2x+1)

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Empty
 one year ago
Best ResponseYou've already chosen the best response.4Yeah, 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
 one year ago
Best ResponseYou've already chosen the best response.0omg thank you so much i understand it now

Empty
 one year ago
Best ResponseYou've already chosen the best response.4Technically you're not allowed to multiply by "dx" like that, but it won't really be a problem until you get to multivariable calculus.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0After 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
 one year ago
Best ResponseYou've already chosen the best response.0similarly, 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
 one year ago
Best ResponseYou've already chosen the best response.0that is cool stuff to know this will help out alot

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0it is called "advanced guessing", becomes easy with some practice

Empty
 one year ago
Best ResponseYou've already chosen the best response.4Is that really called that or are you just joking around XD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Haha here is my theory : guessing \(\cos x\) is an antiderivative of \(\sin x\) is just "plain guessing"

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0but guessing \(e^{2x}\) is an antiderivative of \(2e^{2x}\) is "advanced guessing" as we're jumping ahead avoiding u substitution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is awesome by that theory i have integral of 1/(2x+1)^2 = 1/(2x+1) that is sooo awesome

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0i do remember reading that phrase "advanced guessing" from some textbook :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Yes! 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
 one year ago
Best ResponseYou've already chosen the best response.0but thats only the first step, to be sure, differentiate `1/(2x+1) ` and see if you really get ` 1/(2x+1)^2`

Empty
 one year ago
Best ResponseYou've already chosen the best response.4Yeah, 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 winwin situation haha.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok thats right i see that is important i need to keep that in mind that checking the derivative at the end is important

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you guys for all the help, i gotta get to bed
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