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

1. 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.

2. anonymous

omg thank you so much i understand it now

3. 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.

4. 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 $$\dfrac{1}{x}$$

5. 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 $$\dfrac{1}{x+1}$$

6. anonymous

that is cool stuff to know this will help out alot

7. ganeshie8

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

8. Empty

9. Empty

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

10. anonymous

thats funny

11. ganeshie8

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

12. ganeshie8

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

13. anonymous

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

14. ganeshie8

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

15. 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)

16. ganeshie8

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

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

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

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

20. ganeshie8

gnite!