# I hate Integration by parts

1. Dec 12, 2004

### 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

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

2. Dec 12, 2004

### 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

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

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

Last edited: Dec 12, 2004
3. Dec 12, 2004

### 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

Last edited: Dec 12, 2004
4. Dec 12, 2004

### Pyrrhus

Hey tom,

For

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

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

Last edited: Dec 12, 2004
5. Dec 12, 2004

### quasar987

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

6. Dec 12, 2004

### Tom McCurdy

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

7. Dec 12, 2004

### Tom McCurdy

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

8. Dec 12, 2004

### 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.

9. Dec 12, 2004

### Pyrrhus

That's an integration by parts

If we rearrange it:

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

10. Dec 12, 2004

### quasar987

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

11. Dec 12, 2004

### Pyrrhus

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

12. Dec 12, 2004

### Tom McCurdy

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

Last edited: Dec 12, 2004
13. Dec 12, 2004

### 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

14. Dec 12, 2004

### Pyrrhus

I recommend $u = arctg$

15. Dec 12, 2004

### 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

16. Dec 12, 2004

### 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?

17. Dec 12, 2004

### Pyrrhus

Assuming you mean 5x instead of 52.

Integrate by part that integral

18. Dec 12, 2004

### 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

19. Dec 12, 2004

### Tom McCurdy

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

20. Dec 12, 2004

### 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...