# Integral with substitution

1. Aug 9, 2011

### Shannabel

1. The problem statement, all variables and given/known data
find the integral from 1/3 to 1/2 of 1/x(x+1)(x-1)(x^2+1)

2. Relevant equations

3. The attempt at a solution
expanded and simplified:
1/x^5-x

it gives a hint suggesting i use a substitution here, but i can't find an appropriate one.. Help?

2. Aug 9, 2011

### I like Serena

Welcome to PF, Shannabel!

Perhaps try for instance $u=x^4-1$?

3. Aug 9, 2011

### GreenPrint

Are there suppose to be terms in the numerator besides 1? From what you wrote I would assume no but I wanted to make sure...

4. Aug 9, 2011

### Shannabel

nope, just a 1

5. Aug 9, 2011

### I like Serena

Let me clean that up:
$$\int_{1/3}^{1/2} {1 \over x(x+1)(x-1)(x^2+1)} dx$$

Click the Quote button to see what I did (it is called LaTeX).

6. Aug 9, 2011

### Shannabel

i don't see how that will work.. since then du = 4x^3dx

7. Aug 9, 2011

### GreenPrint

Ya then I would stick with I like Serena's suggestion if you haven't learned partial fraction decomposition yet. Expanding is kind of tedious and there are some very common integrals than can be evaluated once you resolve that into partial fractions.

8. Aug 9, 2011

### I like Serena

Yep. Keep going!

9. Aug 9, 2011

### Shannabel

then i get
du/4x^4(u) ... but i can't integrate that?

10. Aug 9, 2011

### I like Serena

Well, you still have x in your expression.
x should be eliminated... what would you get then?

11. Aug 9, 2011

### Shannabel

how do i eliminate the x?

12. Aug 9, 2011

### I like Serena

Can you solve x from u=x4-1?
And substitute that?

13. Aug 9, 2011

### Shannabel

do you mean like x^4 = u+1?

14. Aug 9, 2011

### I like Serena

Yep!

15. Aug 9, 2011

### Shannabel

then i get
du/u^2+u from 1/3 to 1/2
which is lnu^2 + ln u from 1/3 to 1/2
=2lnu+lnu
=3lnu from 1/3 to 1/2

u = x^4-1
so when x = (1/2), u = -15/16
and when x = (1/3), u = -80/81

so from above i have
3ln(-15/16)-3ln(-80/81)

where did i go wrong? lol

16. Aug 9, 2011

### Shannabel

then i get
du/u^2+u from 1/3 to 1/2
which is lnu^2 + ln u from 1/3 to 1/2
=2lnu+lnu
=3lnu from 1/3 to 1/2

u = x^4-1
so when x = (1/2), u = -15/16
and when x = (1/3), u = -80/81

so from above i have
3ln(-15/16)-3ln(-80/81)

where did i go wrong? lol

17. Aug 9, 2011

### I like Serena

Not so fast!

And I'd really like some parentheses here, otherwise it becomes very confusing.

and now you're saying this integrates to: ln(u^2) + ln u

Let's check that...
Could you perhaps calculate the derivative of that last expression?

18. Aug 9, 2011

### Shannabel

ohhh
(du/u^2)+(du/u)
which integrates to
-1/u + ln u?

19. Aug 9, 2011

### I like Serena

Hmm, let's take a step back. :uhh: