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

Question

empty · Answer

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 · Answer

omg thank you so much i understand it now

empty · Answer

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.

ganeshie8 · Answer

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 · Answer

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 · Answer

that is cool stuff to know this will help out alot

ganeshie8 · Answer

it is called "advanced guessing", becomes easy with some practice

empty · Answer

Lol "advanced guessing"

empty · Answer

Is that really called that or are you just joking around XD

anonymous · Answer

thats funny

ganeshie8 · Answer

Haha here is my theory : 

guessing $\cos x$ is an antiderivative of $\sin x$ is just "plain guessing"

ganeshie8 · Answer

but guessing $e^{2x}$ is an antiderivative of $2e^{2x}$ is "advanced guessing" as we're jumping ahead avoiding u substitution

anonymous · Answer

that is awesome by that theory i have integral of 1/(2x+1)^2 = -1/(2x+1) that is sooo awesome

ganeshie8 · Answer

i do remember reading that phrase "advanced guessing" from some textbook :)

ganeshie8 · Answer

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 · Answer

but thats only the first step,
to be sure, differentiate `-1/(2x+1) `  and see if you really get ` 1/(2x+1)^2`

empty · Answer

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 · Answer

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 · Answer

thank you guys for all the help, i gotta get to bed

ganeshie8 · Answer

gnite!