Question

Rationalizing. Evaluate the integral

Answer 1

\[\int\limits \frac{ 1 }{ 6x+5\sqrt{x} }dx\]

Answer 2

i chose u = 6x+5. du = 1/6du=dx. and x = (u-5)/6. i plugged those in the equation. not sure where to go from here

Answer 3

with my x plugged in i get a big square root in the denominator I'm not sure what to do with it

Answer 4

Woahhh woah woah woah :O
You can't choose u = 6x+5.
I mean you can... but it's NOT going to work out the way you would like.

The 5 is connected to the sqrt(x),

You don't have (6x+5)*sqrt(x).

Answer 5

oh ya you're right! i didn't realize that. what would be a good u?

Answer 6

hmm thinking :d

Answer 7

thank you! Ive been stuck on this for a while

Answer 8

Oh oh oh, ok here's an idea.

Answer 9

Factor a sqrt(x) out of each term in the denominator,\[\large\rm \int\limits\frac{1}{\sqrt{x}~(6\sqrt x+5)}dx\]

Answer 10

Then let \(\large\rm u=6\sqrt{x}+5\)
Give that a shot :) lemme know if you get stuck.

Answer 11

ok cool! thanks!

Answer 12

Here is the intuition behind that choice (just in case it seems random),
I know that the `derivative of square root` is `one over two square roots`.
So I know I can get from sqrt(x) to the 1/sqrt(x) that we factored out somehow by derivative.

Answer 13

I'm actually stuck on the derivative of 6sqrt(x)+5... would the du be 1/6sqrt(x) ?

Answer 14

Do you remember the shortcut for derivative of sqrt(x)?
Yes, you could rewrite it as x^(1/2) and apply power rule,
but square root shows up so often that it's worth remembering.\[\large\rm \frac{d}{dx}\sqrt x=\frac{1}{2\sqrt x}\]

Answer 15

oh ya! so would we multiply 6 to that?

Answer 16

so i get 3/sqrt(x)

Answer 17

Mmm yes, that sounds right!\[\large\rm u=6\sqrt x+5\qquad\to\qquad du=6\frac{1}{2\sqrt x}dx\]

Answer 18

Headed in the right direction now :) yay

Answer 19

yay thanks! i will continue

Answer 20

so i took out 1/sqrt(x) from the integral and 3/sqrt(x) and put them behind the integral. i am left with 1/u inside my integral. is that correct so far ? @zepdrix

Answer 21

You can't take 1/sqrt(x) OUT OF the integral, since the integral depends on x.
You can't take `x stuff` out of the integral.

You can, however, do this\[\large\rm \int\limits\limits\frac{1}{\sqrt{x}~(6\sqrt x+5)}dx\quad=\quad \int\limits\limits\frac{1}{6\sqrt x+5}\left(\frac{1}{\sqrt x}dx\right)\]Unless that's what you were referring to :)

Answer 22

oh thanks for letting me know! i forgot i couldn't do that. so instead, i can pull it out and multiply it as part of dx

Answer 23

would my dx change to sqrt(x)/3?

Answer 24

The only reason to do that is because umm..
when I look at this integral,\[\large\rm \int\limits\frac{1}{\sqrt{x}~(\color{orangered}{6\sqrt x+5})}dx\]I'm thinking, "Ok that orange stuff is my u, so the rest of it is probably my du, let's group it with the dx and check".

Obviously you've had problem lately in which this is NOT the case,
where you have to fill in the extra piece like an x^2,
but sometimes it works to think that way.

Answer 25

\[\large\rm =\int\limits\limits\limits\frac{1}{\color{orangered}{6\sqrt x+5}}\left(\color{royalblue}{\frac{1}{\sqrt x}dx}\right)\]And so far we have,\[\large\rm \color{orangered}{u=6\sqrt x+5},\qquad\qquad du=3\color{royalblue}{\frac{1}{\sqrt x}dx}\]Hmm, so ...

Answer 26

Think of the entire blue clump as your "variable" that you're trying to isolate.

Answer 27

\[\large\rm =\int\limits\limits\limits\limits\frac{1}{\color{orangered}{6\sqrt x+5}}\left(\color{royalblue}{stuff}\right)\]And right now we have:\[\large\rm du=3\color{royalblue}{stuff}\]So how do we solve for the stuff?

Answer 28

we divide? and put it on the other side of du to have dx alone?

Answer 29

so it would be 1/(3/sqrt(x)) which is equal to after multiplying by the reciprocal to sqrt(x)/3 ?

Answer 30

Well we're not trying to get dx alone, right?
We're trying to get `1/sqrt(x) dx` alone, the entire blue thing.
But yes, divide by 3 to isolate it.\[\large\rm \frac{1}{3}du=\color{royalblue}{stuff}\]

Answer 31

hmm ok ... can this go behind my integral?

Answer 32

What do you mean by behind? :o

Answer 33

so it would be \[1/3\int\limits (stuff)\]

Answer 34

so i don't have to deal with it pretty much lol

Answer 35

and just multiply it in the end

Answer 36

but I'm sure it can't be that easy to get rid of it lol

Answer 37

No, it doesn't just get tossed in front of the integral, not yet at least.
This is going to fill in one of the pieces of your integral:\[\large\rm =\int\limits\limits\limits\limits\limits\limits\frac{1}{\color{orangered}{6\sqrt x+5}}\left(\color{royalblue}{stuff}\right)\quad=\quad =\int\limits\limits\limits\limits\limits\limits\frac{1}{\color{orangered}{6\sqrt x+5}}\left(\color{royalblue}{\frac{1}{3}du}\right)\]Like this, ya? :d

Answer 38

But yes, after you've correctly plugged it in, you can drag the 1/3 out front :)

Answer 39

yes ok thanks! i will try it

Answer 40

ohhh do i get 1/3 because I'm multiplying 1/sqrt(x) * sqrt(x)/3 therefore the sqrt(x)'s cancel out and i get 1/3 ! ?

Answer 41

and then i can put the 1/3 behind the integral?

Answer 42

Oh boy you're not getting it :U
Let's try this again hehe.\[\large\rm du=3\frac{1}{\sqrt x}dx\] `You're not solving for dx`.

Answer 43

`You're solving for 1/sqrt x dx`

Answer 44

Which means you should only be moving the 3.
Nothing else.

Answer 45

No? :o
Brain esplode?

Answer 46

lol sorry >.< moving the 3 behind the integral?

Answer 47

No.
I'm saying that, at no point should be you multiplying anything by sqrt(x)/3.
That is not a step we want to apply.

Answer 48

but sqrt(x)/3 is our dx right?

Answer 49

\[\large\rm \frac{\sqrt x}{3}du=\frac{\sqrt x}{3}\frac{3}{\sqrt x}dx\]This solves for dx,\[\large\rm \frac{\sqrt x}{3}du=dx\]Bad.
That's not what we want.

Answer 50

ohh ok

Answer 51

so what i should focus on is finding the integral of 1/6(sqrt(X)+5

Answer 52

and leave the dx alone for now

Answer 53

\[\large\rm du=3\frac{1}{\sqrt x}dx\]Dividing by 3,\[\large\rm \frac{1}{3}du=\left(\frac{1}{\sqrt x}dx\right)\]Good.
This is what we want.
Notice that the right side is what we have in our integral.

No, don't leave the dx alone.
`You have to focus on plugging in the pieces correctly` :)

Answer 54

ohh ok i see!

Answer 55

We did some fancy business and agreed that this:\[\large\rm \int\limits\frac{1}{6x+5\sqrt{x}}dx\]is equivalent to this:\[\large\rm \int\limits\frac{1}{\color{orangered}{6\sqrt x+5}}\color{royalblue}{\left(\frac{1}{\sqrt x}dx\right)}\]Pay attention to the colors.

And we decided to let our \(\large\rm \color{orangered}{u=6\sqrt x+5}\)
Taking the derivative, and then dividing by 3 gave us this: \(\large\rm \color{royalblue}{\frac{1}{3}du=\left(\frac{1}{\sqrt{x}}dx\right)}\)

Answer 56

so now i have 1/u multiplied to 1/sqrt(x) dx.... i hope this is correct :x

Answer 57

Plug in both pieces,\[\large\rm \int\limits\limits\frac{1}{\color{orangered}{6\sqrt x+5}}\color{royalblue}{\left(\frac{1}{\sqrt x}dx\right)}\quad=\quad \int\limits\limits\frac{1}{\color{orangered}{u}}\color{royalblue}{\left(\frac{1}{3}du\right)}\]not just the u.

Answer 58

oh yes ok!

Answer 59

so i can have the integral of 1/3u

Answer 60

But let's pull the 1/3 out front like you had suggested before :)\[\large\rm =\frac{1}{3}\int\limits \frac{1}{u}du\]

Answer 61

wow ok yes that makes it easy to integrate

Answer 62

thanks for your patience lol i see it now! so the integral of that would be ln u ?

Answer 63

\[\large\rm =\frac{1}{3}\ln|u|+c\]Mmmm ok sounds good so far.

Answer 64

1/3 * ln(u) and plug u back in

Answer 65

yay ok! 1/3ln|6sqrt(x) + 5| + c

Answer 66

thanks so much for your help!!!

Answer 67

And notice that since sqrt(x) is always positive,
6sqrt(x)+5 is always positive.
So we don't actually need the absolute bars giving us the positive of this argument.\[\large\rm =\frac{1}{3}\ln\left(6\sqrt x+5\right)+c\]

Yayyy team! ୧ʕ•̀ᴥ•́ʔ୨

Answer 68

oh true! yay thank you!!!!!!!!!