Question

calculus @zepdrix

Answer 1

Would the antiderivative of \(\Large \frac{x^{-1}}{2}\) equal \(\large \ln(\frac{1}{2}x)\) or \(\large\frac{1}{2}\ln x\)? Or neither...
@zepdrix @ParthKohli

Answer 2

\[\large\rm \int\limits \frac{1}{2x}dx\quad=\quad \frac{1}{2}\cdot\int\limits \frac{1}{x}dx\quad=\quad \frac{1}{2}\cdot \ln|x|\]

Answer 3

How come you can pull out the constant like that?

Answer 4

Because constant coefficients are stupid,
nobody loves them.

Answer 5

Ummm

Answer 6

lol, ok

Answer 7

It's the same with limits and such,
hopefully you remember back to those ideas...

The only things you can't pull out,
are the things that are changing,
so anything involving x.

1/2 does not ever change.
So the integral doesn't depend on it in any way.

Answer 8

ok

Answer 9

Just be careful how you pull them out,
realize when you pull it out front,
it now represents a coefficient multiplying THE ENTIRE THING.

Example:\[\large\rm \int\limits 2x+2~dx\quad\ne\quad 2\int\limits x+2~dx\]That 2 is multiplying the entire thing!
Which would give us 2x+4 if we put it back in.

Answer 10

so would this be correct?

Answer 11

\[\large\rm \int\limits 2x+2~dx\quad=\quad \int\limits2(x+1)dx\quad=\quad 2\int\limits x+1~dx\]

Answer 12

No no no,\[\large\rm (1+2x)^{-1}\quad\ne\quad 1^{-1}+(2x)^{-1}\]In general,\[\large\rm (x+y)^a\quad\ne\quad x^a+y^a\]

Answer 13

Oh...okay, so what do I do?

Answer 14

Well, this integral is a tad trickier than the ones you've dealt with before.
You can either do a quick substitution, u=1+2x,
OR
preferably,
you can just sort of guess your way along.
There is this technique called "advanced guessing" that we use often for problems like this.

Answer 15

We know that:\[\large\rm \int\limits\frac{1}{x}dx\quad=\quad \ln|x|\]So we're expecting that \(\large\rm\int\dfrac{1}{1+2x}dx\) should give us "something like" \(\large\rm ln|1+2x|\) right?

Answer 16

yeah. I did the u substitution and got \(\Large \frac{1}{2}\ln|1+2x| +C\) is that correct?

Answer 17

Yay good job \c:/

Answer 18

Thanks :)

Answer 19

u-sub is really burdensome for a problem like this,
errr i should say, it will be later on.

But I suppose at this stage it's good to practice :))

Answer 20

hey because derivatie and integrals can be split into term by term

Answer 21

d/dx(f+g)= df/dx + dg/dx

Answer 22

we can go into more detail why the derivative is associative if u would like to learn

Answer 23

we can go into more detail why the derivative is commutative if u would like to learn

Answer 24

we can go into more detail why the derivative is distributive if u would like to learn

Answer 25

lol thats the right one