I hate 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 :smile:

[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. :biggrin:

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} ....