What is the best approach for Integration by parts?

[Tom McCurdy]

I am usually alright once I figure out how to split up the integral into
u:
du:
v:
dv:
so i can simply do
$$uv-\int v*du$$ but I keep messing up on there I will post some examples if I can find them and if someone could help me that would be great

$$\int (ln(x))^2$$

 

[Tom McCurdy]

Here is what I am trying

$$\int (ln(x))^2$$

u: ln x
du: 1/x
v: x ln(x)-x
dv: ln x

$$(x ln(x)-x)ln(x)-\int {ln(x)}-1$$

$$(x ln(x)-x)ln(x)-(x ln(x) -x) + x$$

 

[Tom McCurdy]

I am also trying to figure out u and dv for $$\int (x ln x)^2$$ but i figured it would follow in line once i figured out the previous problem

 

[Pyrrhus]

Hey tom,

For

$$\int x^2 (\ln x)^2 dx$$

Use $u = (\ln x)^2$

 

[quasar987]

Here is what I am trying

$$\int (ln(x))^2$$

u: ln x
du: 1/x
v: x ln(x)-x
dv: ln x

$$(x ln(x)-x)ln(x)-\int {ln(x)}-1$$

$$(x ln(x)-x)ln(x)-(x ln(x) -x) + x$$

You have the right answer, what's the problem?

 

[Tom McCurdy]

You have the right answer, what's the problem?

I figured it out and kept editing it... thanks though... I am still not sure how I would know to split up the U and dv in the way that I did

 

[Tom McCurdy]

ok I got it down to
$$(x^2 ln(x))(x ln(x) - x) - [ \frac{x^3(3ln(x)-1)}{27}-\frac{x^3}{27} ]$$

 

[Tom McCurdy]

out of curisoity is there any rules on how to do $$\int \frac{x^2 ln(x)}{27}$$ because it was needed for the previous problem and I just used my calculator to come out with the first fraction after the first bracket.

 

[Pyrrhus]

That's an integration by parts

If we rearrange it:

$$\frac{1}{27} \int x^2 \ln x dx$$

 

[quasar987]

I am still not sure how I would know to split up the U and dv in the way that I did

Unfortunately, I don't think there's any other way than trial an error.

 

[Pyrrhus]

Quasar is right, but you will learn to notice which to pick as you make plenty integration by parts.

 

[Tom McCurdy]

thx... what would I do to start $$\int tan^{-1}$$

 

[Tom McCurdy]

oh wait the derivitive of $$tan^{-1} = \frac{1}{(x^2+1)}$$ So I set tan^-1 as u and 1 as dv... let me see what i get

 

[Pyrrhus]

Assuming your mean arctg

I recommend $u = arctg$

 

[Tom McCurdy]

I was doing the $$\int x sin^{-1}$$ and was wondering once you got to
$$x sin^{-1} - \int \frac{x}{\sqrt{1-x^2}}$$ how you would integrate $$\int \frac{x}{\sqrt{1-x^2}}$$ I know the answer is $$-\sqrt{1-x^2}$$ but how do you get that

 

[Tom McCurdy]

Alright
I am doing the $$\int x^2 e^{5x}$$ and I got to here
assuming u:x^2 and dv: e^(5x)
$$\frac{x^2*e^{5x}}{5} - \frac{2}{5} \int 2x e^{52}$$

what would I do now?

 

[Pyrrhus]

Assuming you mean 5x instead of 52.

Integrate by part that integral

 

[Tom McCurdy]

oh ****... this is going to take forever... I am on problem 18 out of 64 and I realized that the problems ahead will require 5 integration by parts

 

[Tom McCurdy]

lol this is what i mean $$\int x^5*e^{2x^3}$$

 

[Tom McCurdy]

woah this was werid...

i was doing $$\int x^3*e^{x^2}$$ and I got down to

$$2x^4 e^{x^2}-2 \int x^3*e^{x^2}$$ which is the original problem... how would I go about solving this...

 

[Tom McCurdy]

Since

$$\int x^3*e^{x^2} = 2x^4 e^{x^2}-2 \int x^3*e^{x^2}$$ can I just say
$$3 \int x^3*e^{x^2} = 2x^4 e^{x^2}$$
$$\int x^3*e^{x^2} = \frac{2x^4 e^{x^2}}{3}$$ ?

 

[Pyrrhus]

Yes you can.

 

[NeutronStar]

It's True! Lazy People Ride The Elevator!

I was always taught to pick $$u$$ by the following priority,…

Inverse Trig
Logarithms
Polynomials
Rational
Trigonometric
Exponential'

The order can be remembered by the saying at the top of this post.

In the case that you were working with here you should let $$u$$ equal the natural log function and then let $$dv$$ equal what's left.

This system may not be fail-proof, but it seems to work most of the time. Actually it's never failed me yet.

 

[Tom McCurdy]

thx... I am having trouble with $$\int x^5*e^{2x^3}$$

what should be U and what should be dv...

I choose u:x^5 and dv: e^(2x^3) but it got out of hand quickly

 

[Tom McCurdy]

oh shoot... I see a mistake I made... I had it as
$$\int x^5*e^{{(2x)}^3}$$
instead of
$$\int x^5*e^{2x^3}$$

 

[Tom McCurdy]

nope still have a problem when I use u: x^5 and dv: e^(2x^3)
the power increases and after first one you get

$$6x^7 e^{2x^3} - 30 \int x^2 e^{2x^3}$$... what should i do

 

[Pyrrhus]

Keep on integrating, and pray you didn't make a mistake

Also a note: I've noticed you don't like putting dx in your integrals, do it on your work, some teachers (specially mathematicians) take points off if you don't.

 

[Tom McCurdy]

nope still have a problem when I use u: x^5 and dv: e^(2x^3)
the power increases and after first one you get

$$6x^7 e^{2x^3} - 30 \int x^2 e^{2x^3}$$... what should i do

sorry i mistyped last time
the reason i am nervous that i did something wrong is that the power inside the integral increased instead of decreased
it went from 5 to 6
correct version
$$6x^7 e^{2x^3} - 30 \int x^6e^{2x^3}$$...