Integration of fractions???!

[slow ledge]

What have you tried so far?

[idle bloom]

the thing is i always tend to bring the fraction up

but i know thats not how im supposed to do it

[slow ledge]

What do you mean, bring the fraction up?

[idle bloom]

3x(5-x^2)^-1

[slow ledge]

Well, that doesn't hurt, but that doesn't help.

[idle bloom]

how do i tackle these questions?

[slow ledge]

The technique I'd use here is substitution

[idle bloom]

could u teach me?

i saw it on youtube but i didnt quite understand it

[slow ledge]

Essentially, it consists of taking part of the integrand, and assigning it a variable, so, let's try a simpler example to make sure you understand it

[idle bloom]

ok!

this is my weakest component in integrating so if i can master this it would be really great

[slow ledge]

It's not particularly difficult, I'm sure you'll get it in no time 🙃

So, say you had $\int \frac 1{x-1}\dd x$

[past warren]

𝔧𝔪𝓣𝛾𝜑𝜽

[idle bloom]

yes

ln|x-1| + c

[slow ledge]

Did you just memorize the anti derivative of that function?

[idle bloom]

i know that 1/ax+b = 1/a ln|ax+b|

[slow ledge]

Ah, I see. Well, I wanted to use that integral to show you substitution

[idle bloom]

thats ok please continue

[slow ledge]

in this case, if you don't know that formula, since you can't integrate $(x-1)^-1$ directly, we perform a substitution. It might look like this:
$\\text{Let } u = x-1\\dv{x}(u) = \dv{x}(x-1)\\dv{u}{x} = 1 \ \dd u = \dd x$

oh god

[idle bloom]

erm

[slow ledge]

Yeah let me fix that up 💀

[idle bloom]

how do u get from step 1 to 2?

[past warren]

𝔧𝔪𝓣𝛾𝜑𝜽

[idle bloom]

why is it keep getting deleted

[slow ledge]

I differentiated both sides

[slow ledge]

Cause I edit it. I made mistakes 💀

[idle bloom]

oh hah

so i let the denomiator = u

[slow ledge]

Right, but then, we need to find what dx becomes after the transformation, so we differentiate both sides and isolate it

[idle bloom]

then i differentiate u?

[slow ledge]

Exactly

[idle bloom]

which is 1

then i differentiate x-1

=1

[slow ledge]

Wait

You differentiate u with respect to x, don't forget u is a function of x. So, d/dx (u) = u' = du/dx

[idle bloom]

wut

[slow ledge]

Since we said u = x-1, we know u is a function of x. If I told you, what's the derivative of f(x) with respect to x, what would you say

[idle bloom]

1

[slow ledge]

(here f(x) is just a random function)

[idle bloom]

idk

[slow ledge]

it would be f'(x)

[idle bloom]

what is '

[slow ledge]

oh, have you not seen that notation for derivatives?

it means take the derivative of f

[idle bloom]

i did but i forgot

[slow ledge]

I see

well, another way to write f'(x), is df/dx, in other words, the derivative of f, with respect to x

And that's precisely what we're doing, just replace f with u

[idle bloom]

i dont know how to end up with du=dx

[slow ledge]

Do you understand up until du/dx = 1

[idle bloom]

dont u just differentiate both of them

[slow ledge]

That's the second step here, yes

[idle bloom]

so i differentiate u i get 1 i differentiate x-1 i get 1

is that correct?

[slow ledge]

if we differentiated u with respect to u, yes, but we're differentiating it with respect to x

[idle bloom]

then its 0

cus its a constant

[slow ledge]

That's the thing, u isn't a constant, it's a function of x. remember, u = x-1

[idle bloom]

uh huh

[idle bloom]

yeah i get step 2 now, what about step 3

du/dx =1

[slow ledge]

du/dx is notation that means "the derivative of u with respect to x"

[idle bloom]

so how does that work?

[slow ledge]

I'm not super sure how to explain it, maybe you need a little revision on derivatives. If I can suggest a youtube series (maybe not the whole thing, but you can see which episodes are useful to you), it would be "The essence of Calculus" by 3Blue1Brown. It explains limits, derivatives, and integrals really well with visuals

[idle bloom]

i dont think i can afford to watch a video rn though... my exam is in 2 days 😦

could u explain it? it doesnt have to be super good... i just needa be able to understand it

@slow ledge please?

[stiff prairie]

Brother

Just learn substitution

[idle bloom]

im trying to understand it here

[stiff prairie]

what do you not understand

show me

[idle bloom]

so i let the denomiator = u correct?

[stiff prairie]

no

keep it simple

[idle bloom]

wut

[stiff prairie]

u = the part that has the higher power

[rich sigil]

@idle bloom you could just professor Leonard in youtube channel to understand integration

[idle bloom]

how would u integrate this?

[stiff prairie]

between 3x and x^2 which has the higher power

?

[idle bloom]

x^2

[stiff prairie]

yes

so substitute the part that has the x^2 in it

let u = that

then find du

[idle bloom]

so the denominator

[stiff prairie]

yes

[idle bloom]

i differentiate in terms of u>

or wut

[stiff prairie]

yeah

d for derivative

[idle bloom]

isnt that just 1

[stiff prairie]

is what 1?

[idle bloom]

wait let me be clear

for this question u= 5-x^2

[stiff prairie]

exactly

so differentiate that

[idle bloom]

then i differentiate this expression

2x

[idle bloom]

-2x

[stiff prairie]

-2x

[idle bloom]

yeah i dont get that

[stiff prairie]

ignore it

we're talking it out

so what does du =?

[idle bloom]

2x?

[stiff prairie]

sorry im just trying to help figure what he doesnt understand

nu

-2xdx

[unreal arrow]

got it your point

[idle bloom]

?

im getting confused

[rich sigil]

@idle bloom do you know precalculus

[idle bloom]

whats that? example?

[rich sigil]

Like path between algebra and calculus

[stiff prairie]

du/dx is a derivate its also a fraction

[stiff prairie]

nvm

[rich sigil]

Like things we need to know before diving into calculus

[idle bloom]

differentiate the function x in terms of u

[rich sigil]

@idle bloom do you know algebra 2?

[idle bloom]

whats that? give examples

[rich sigil]

;-;

[stiff prairie]

You need to learn derivate notation

[rich sigil]

Like lograthim, natural lograthim, complex numbers, complex functions

[idle bloom]

yea

[hot steeple]

most times they're substitution or fit in a trig integral

[idle bloom]

can anyone do the question and explain it to me?

[unreal arrow]

i have best method for you

just start from scratch

we have 3x/5-x^2

[idle bloom]

yes

[unreal arrow]

now there is power 2

[idle bloom]

yes

[unreal arrow]

now we need multiple of 2 in coefficient

ok

[idle bloom]

???

[unreal arrow]

ok you are confusing here

[stiff prairie]

HE DOESNT KNOW DERIVATIVE NOTATION

[idle bloom]

yeah i dont

[unreal arrow]

just divert our root

[stiff prairie]

u = x^2

differentiate this

[idle bloom]

2x

[stiff prairie]

let me rewrite it

[unreal arrow]

just diff. whole term 5-x^2

[stiff prairie]

f(x)=x^2

[unreal arrow]

make it more easy

[stiff prairie]

find f'(x)

[candid sparrow]

did

d

did

s

s

he

def

fes

s

s

dr

[idle bloom]

?

[candid sparrow]

rf

sorry

my brother did that

sorry

[idle bloom]

is that symbol mean integrate? or wut

[rich sigil]

2x

[unreal arrow]

f means function

[idle bloom]

'

[stiff prairie]

f' is read f prime

[unreal arrow]

this is not a sign of inte.

[idle bloom]

im so dumb 😭

[idle bloom]

ok continue...

[stiff prairie]

it means the first derivative of the function f

[idle bloom]

so it just means differentiated

[rich sigil]

I suggest you to read aops book on algebra, which is currently im doing for imo

[stiff prairie]

yes

[idle bloom]

ok

so 5-x^2= -2x

then?

[unreal arrow]

we let 5-x^2=t

and diff. wrt x

we get -2x dx=dt

we already have x dx in our fun.

so we can write it as x dx = (-1/2)dt

[idle bloom]

holy this is confusing

[unreal arrow]

on which part you are confusing

[idle bloom]

is there another way instead of subsitition

[unreal arrow]

we have but that may be more lanthy

int{u.v}dx

[unreal arrow]

know about this

[stiff prairie]

xd

[fiery swift]

Has this been solved yet? .o.

@idle bloom ^

[idle bloom]

no 😦

[fiery swift]

Oh dang

Okay I can try to explain it to you!

Hopefully that works out some kinks

I will be a bit busy for the next 2 hours but send question in here if you have any!

[rich sigil]

Legends say that he still searching to solve the problem

[idle bloom]

oh interesting

[idle bloom]

and yes

i sound so stupid when i do things like integration and differentiation

@white reef

here!

[white reef]

do you know u substitution

[idle bloom]

i dont know

we never learnt

[white reef]

where did you get this problem from?

[idle bloom]

from a exam paper

but it was never in my notes 🥹

[white reef]

huh okay

well I can't think of a way to do this other than u sub

[idle bloom]

my friend told me to take out -3/2

but idk how he knows that

[white reef]

so the way u sub works is that we change from the integral being over x to being over u

so what we can do is we can set $5 - x^2 = u$

[past warren]

Ninja Duck

[idle bloom]

ok i got that part

[white reef]

we also need to change the $\dd{x}$

[past warren]

Ninja Duck

[white reef]

we know $\dv{u}{x} = -2x$ right?

[past warren]

Ninja Duck

[idle bloom]

yes

[white reef]

so $\dd{x} = -\frac{\dd{u}}{2x}$

[past warren]

Ninja Duck

[idle bloom]

yes cross multiply

[white reef]

now your integral becomes $\int_{-1}^0 \frac{3x}{5-x^2} \dd{x} = \int_?^? \frac{3x}{u} \frac{\dd{u}}{-2x}$

now notice that the limits of integration also change, since earlier it was for where x varies across, but now we have u

[idle bloom]

wait

[white reef]

i havent calculated what the new limits are yet so for now im leaving them as question marks

we will find them in a second

[white reef]

yes?

[idle bloom]

i understand until here

[white reef]

it's -du / 2x

not just 2

[idle bloom]

oh its not the same thing?

[white reef]

why would it be the same thing?

[idle bloom]

oops i forgot the x haha

[white reef]

yeah

[idle bloom]

ok then after that what

[white reef]

now the denominator us u, so im replacing that with u

and dx is -1/2x du so im replacing dx with that

also there should be a minus sign

[idle bloom]

what about the negative sign?

[white reef]

yeah i forgor

now we can cancel the xs

[idle bloom]

erm wut

[white reef]

so you get $-\int_?^? \frac{3}{2u}\dd{u}$

[past warren]

Ninja Duck

[white reef]

what's the problem?

[idle bloom]

dont get the last part

[past warren]

Ninja Duck

[white reef]

does this make more sense?

[idle bloom]

yes

[white reef]

you can see how this is that same as what i wrote earlier right?

i guess i just did one too many steps at once

[idle bloom]

yes

[white reef]

and do you see how it cancels to this?

[idle bloom]

yes

[white reef]

now we have an integral only in terms of u

with a du

so we can almost just integrate this

but we don't know the limits

the old limits used to be
from x = -2 to x = 0

[idle bloom]

why do limits change?

[white reef]

because u isn't x

[idle bloom]

ah ok

[white reef]

so when $x = -2$

$u = 5 - x^2 = 5 - (-2)^2 = 5 - 4 = 1$

and

when $x = 0$

$u = 5 - x^2 = 5 - (0)^2 = 5$

[past warren]

Ninja Duck

[white reef]

so while x varies from -2 to 0
u varies from 1 to 5

does this make sense?

[idle bloom]

ok i gets

[white reef]

u is a function of x rememeber

[idle bloom]

can u give me a similar problem for me to try?

[white reef]

its like how when t varies from 1 to 3, t^2 varies from 1 to 9

[idle bloom]

yes

i understand

[white reef]

sure $\int_{-1}^2 \frac{2x}{x^2-2} \dd{x}$

[past warren]

Ninja Duck

[idle bloom]

lemme handwrite it

yes?

[white reef]

,ask solve \int_{-1}^2 \frac{2x}{x^2-2}

what

[idle bloom]

i got ln2

[past warren]

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[white reef]

oh huh

oh right oof

yeah i kinda messed up there

[idle bloom]

hmh?

does it look correct?

[white reef]

,rotate

[white reef]

yeah the method is correct

but this isn't technically correct cause if you look at the function

,ask graph 2x/(x^2 - 2)

[past warren]

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Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by officiallyaninja

[white reef]

yeah it goes off to infinity at sqrt 2

[idle bloom]

erm so im wrong?

[white reef]

no it's my fault i gave you a problem without a solution

but the method was right

[idle bloom]

thats ok i just wanted to know if i used the correct method

[white reef]

the method only works if the integral doesn't go to infinity within the limits of integration

[idle bloom]

does this also work when the numerator power is higher than denominator?

no right… cus u cant cancel the x on top to make it an integral

[white reef]

this works in general when the numerator is proportional to the derivative of the denominator

$\int \frac{f'(x)}{f(x)} \dd{x}$

let $u = f(x) \implies \dv{u}{x} = f'(x) \implies \dd{x} = f'(x)\dd{u}$

[past warren]

Ninja Duck

[white reef]

then the integral becomes $\int \frac{1}{u} \dd{u}$

[past warren]

Ninja Duck

[idle bloom]

so same thing…

can i get one more qn? To practice

[white reef]

,ask integral of 5x / (x^2 +1) from 2 to 4

okay this works so try this

wsit

no

[past warren]

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[white reef]

yeah this

see if you get that

[fiery swift]

In more detail ig, what ninja duck is referring to is that when you integrate, we usually see variables of x and the end of an integral being "dx" which tells us what variables we actually are integrating (you won't have to worry about more than one variable rn). The dx corresponds to it's bounds. In this case, they are x bounds. If we switch to du, then we have to convert the old x bounds (-2 and 0) into u bounds! How we do this is simple: we plug in one of the old x bounds--let's say -2 -- into what u equals to get the proper new u-bound of 1 and likewise for the other bound of 0 that, after plugging it back into the u equation (5-x^2), we get the new u-bound of 5. These bounds are exactly replaced in the locations of the old ones, making the new integral have bounds that go from 1 to 5 of a new integrand that deals with "du".

Hopefully this bit of clarification helps

[idle bloom]

can u check what went wrong?

@white reef @fiery swift

i dont got it 😭

[fiery swift]

Lemme see

nice with the pick of the "u"

one sec.

so if u = x^2 + 1, then that implies that du = 2x dx which implies that du/dx is equal to 2x. That part is also correct there, so nice job on that bit

checking further..

the step of dx = du/(2x) is also correct..

so now for the bounds..

If x = 4, (the top bound), then we can plug in this top bound into the u = ... equation

( to get the top bound with respect to "u" )

[idle bloom]

i didnt get the same answer

[fiery swift]

plugging this in, we get u = 4^2 + 1 which is to say that the new integral's top bound is 17

Using the same method, the new lower bound is 5

Okay, here's the tricky part, the new integrand..

ah, I see

[idle bloom]

yes 😭

[fiery swift]

So when we switch from x --> u, we shouldn't see any x's anymore

rather, everything should be in terms of "u", for this problem now

(at this step*)

Well, looking at the previous integrand

[idle bloom]

i did that no?

[fiery swift]

I don't believe so, when you're integrating in terms of du, you shouldn't have a 2x there in the new integrand along with it.

[idle bloom]

i cancelled it from the 5x / 2x

[fiery swift]

Oh btw, do you have VC abilities? I would find it much easier to help with that .o.

[idle bloom]

i cant talk rn 🙈

[fiery swift]

No worries

If you want, I can speak to you and you can type?

but no worries if you don't wanna do that either!

I won't take it personally lol

Just helpin' 👍

[idle bloom]

yeah lets just do text u can do voice memo here

[fiery swift]

voice memo? .o.

[idle bloom]

or in dms

[idle bloom]

like voice record

[fiery swift]

Oh!

okay okay

[idle bloom]

in dms only though

[fiery swift]

Sounds good!

[fiery swift]

Oh yeah don't forget the solved status!

[sinful radish]

.solved