Question

how do you find the anti derivative of :

1) x(2x+1)^2
?? please help!

Answer 1

* without multiplying it all out!

Answer 2

anti derivative = integration
Do you know integration?

Answer 3

u-sub

Answer 4

remember the chain rule, @Laurenn_215

Answer 5

yeah i know integration eccept when theres multiplying involved in integration it gets difficult!

Answer 6

yeah i know the chain rule except i cant use it for this one because of the x infront?

Answer 7

it says, the derivative of f(g(x))=g'(x)f'(g(x))

Answer 8

what if we integrated both sides?

Answer 9

then, f(g(x))=Integration[g'(x)f'(g(x))]

Answer 10

hmmm

Answer 11

* without multiplying it all out! --> why do you want to make your life more difficult ??

Answer 12

And, that's with respect to g(x)

Answer 13

because when you get questions like 45x(3x - 2)^4 you dont want to spend 7000 hours in an exam multiplying it all out haha

Answer 14

Now, I'm going to give you a rule, and explain it.

Answer 15

okay thanks inkyvoyd :)

Answer 16

∫ u'f(u) dx where u is a differentiable function of x
=∫ f(u) du

Answer 17

This equation and the general method of using it for integration is called integration by u-substitution.

Answer 18

Why? Because we subsitute "u" for a function g(x)

Answer 19

Now, let me prove the rule.

Answer 20

*explain it

Answer 21

oh! do you mean the product rule?!

Answer 22

Ok. in that case The problem you just gave know can be sorted out by taking
t = 3x - 2
dt = 3 dx
\[\int\limits_{}^{}\frac{45}{3} \frac{t-2}{3}*t^4dt\]

Answer 23

∫ u'f(u) dx
du/dx=u' -> rewrite in differential form du= u' dx
-> put inside the indefinite integral ∫ u'f(u) dx
you'll get ∫ f(u) du

Answer 24

Now you can multiply it all out

Answer 25

This is actually an analogy of the chain rule.

Answer 26

ohh

Answer 27

∫ u'f(u) dx, where u'=du/dx
∫ u'f(u) dx=∫ f(u)u' dx
rewrite du/dx=u' in differential form
-> du=u' dx
look at ∫ f(u)u' dx and du= u' dx
we can put du in ∫ f(u)u' dx
becomes ∫ f(u) du

Answer 28

Now let's apply this rule to your example.

Answer 29

omg. Mybad. I thoguht that was x^2

Answer 30

That's an x?

Answer 31

@Laurenn_215 , my bad. This is an integration by parts problem.

Answer 32

oh hehe no problem!

Answer 33

@inkvoyd,not eactly

Answer 34

@shivam_bhalla , I can turn it into one

Answer 35

@Laurenn_215 , let's play around with the product rule.

Answer 36

the derivative of f(x)g(x)=f'(x)g(x)+f(x)g'(x) right?

Answer 37

Let's integrate both sides

Answer 38

okay :) yep

Answer 39

take t = 2x+1
dt = 2dx
dt/2 = dx
\[\int\limits_{}^{} \frac{t-1}{2} * t^2 * \frac{dt}{2}\]
NOw multiply it out, integrate and finally substitute t back with 2x+1

Answer 40

@inkyvoyd ,got it?

Answer 41

f(x)g(x) dx=∫ f'(x)g(x) + ∫ f(x)g'(x) dx

Answer 42

yep thats the product rule inkyvod?

Answer 43

@shivam_bhalla , nope, but you u-subbed, so you win. I'm going to explain integrationby parts though

Answer 44

Apply the chain rule and you will get the answer.
\[\int\limits_{?}^{?}f(x)g(x)=f(x)\int\limits_{?}^{?}g(x)dx-\int\limits_{?}^{?}(\int\limits_{?}^{?}g(x)dx)f \prime(x)dx\]
Take f(x)=x and g(x)=\[(2x+1)^{2}\]
It is certain to work.

Answer 45

Looks lke you want to go the long way , so continue...

Answer 46

f(x)g(x) dx=∫ f'(x)g(x) dx + ∫ f(x)g'(x) dx -> let's move stuff around
∫ f(x)g'(x) dx=f(x)g(x) dx-∫ f'(x)g(x) dx

Answer 47

set f(x) to (2x+1)^2 and g'(x) to x

Answer 48

then you have your answer.

Answer 49

Give the medal to @shivam_bhalla , because he gave a better and faster answer.

Answer 50

I only showed you how I would (stupidly) do it because it's a techinque that one might be interested in.

Answer 51

aw okay!

Answer 52

No, give it to @inkvoyd, , you will need the concept of parts later. Learnt it now

Answer 53

wow guys

Answer 54

*learn

Answer 55

uhh... thanks...

Answer 56

@inkyvoyd , you are good at explaining things. A good teacher :D

Answer 57

Thanks :D. Unfortunately, I can't do u-sub for life lol.

Answer 58

Maybe I should *do* those practice problems I have lying around the house

Answer 59

wait, practice. NO NEVER!

Answer 60

*t-sub. lol

Answer 61

LOL, these substitution make your life hell lot easier , so learn it. (just and advice)

Answer 62

*an

Answer 63

I'll try ;)

Answer 64

Sorry about the delay. I was well into an answer and a kind of lock up happened. Here I go again!

Answer 65

The required integral is as follows:
\[x/6(2x+1)^{3}-(2x+1)^{4}/48\]
I will post an explanation of how it is found. Please wait.

Answer 66

The formula for the integral of a product is as follows:
Integral of 1st times 2nd equals1st times integral of 2nd minus integral of (differential coefficient of 1st times integral of 2nd).
I will now post the working out of this.

Answer 67

\[x(2x+1)^{3}/6-\int\limits_{}^{}(2x+1)^{3}/6 \]Take x to be the 1st and take (2x + 1)^2 to be the 2nd.
Then using the formula we get:
\[I=x(2x+1)^{3}/6-\int\limits_{}^{}(2x+1)^{3}/6\]

Answer 68

Next step is to find the integral as follows:
\[\int\limits_{}^{}(2x+1)^{3}/6=(2x+1)^{4}/48\]

Answer 69

So we now have:
\[I=x/6(2x+1)^{3}-(2x+1)^{4}/48+C\]
If you differentiate the above expression you will find the original expression that was required to be integrated.

Answer 70

I hope that you can follow. Please ask if you need any clarification of a step.

Answer 71

The expression for the integral can be simplified down to the following:
\[x ^{4}+4x ^{3}/3+x ^{2}/2+C\]