1. Jan 26, 2008

### BuBbLeS01

1. The problem statement, all variables and given/known data
Find the integral...
2t / (16t^4 + 1)

2. Relevant equations

3. The attempt at a solution
I am stuck on this problem...I have tried long division but that didn't work out for me. I don't know how to solve this.

2. Jan 26, 2008

### rocomath

Long division should be reserved for improper fractions.

$$\int\frac{2t}{(4t^2)^2+1}$$

Can you take it from here?

3. Jan 26, 2008

### BuBbLeS01

oh ok so i can use u substitution
u = 4t^2
du = 8t dt
1/4du = 2t
1/4 ln (4t^2)^2 + 1
is that right?

4. Jan 26, 2008

### rocomath

You're substitutions were right. But that is not correct.

$$\frac 1 4 \int \frac{du}{u^2+1}$$

5. Jan 26, 2008

### BuBbLeS01

On my paper it says hint: use inverse trig functions. But I don't know how to use one of them because they don't resemble that.

6. Jan 26, 2008

### rocomath

Keep searching! It's very common! Sine, Cosine, or Tangent.

7. Jan 26, 2008

### BuBbLeS01

I can't use arctan because on the bottom it is a^2 + u^2 and u has to have the x in it right?

8. Jan 26, 2008

### rocomath

It doesn't matter! You used a substitution and swapped variables.

a, is simply a constant.

$$\int\frac{dx}{x^2+a^2}=\int\frac{du}{u^2+1^2}$$

9. Jan 26, 2008

### BuBbLeS01

oh ok so I have...
1/4 *1/1 arctan |4t^2| / 1 + C
1/4 Arctan |4t^2| + C

10. Jan 26, 2008

### rocomath

Good!

11. Jan 26, 2008

### BuBbLeS01

oh yay thanks!
One more question...
how do you find the integral of something with dx on top like...
dx / (x^2 - 4x + 20)

12. Jan 26, 2008

### rocomath

It doesn't matter what variables you're using, just pay attention to what you're Integrating with respects to and treat the others as constants.

$$\int\frac{dx}{x^2-4x+20}$$

Try completing the square.

Last edited: Jan 26, 2008
13. Jan 26, 2008

### BuBbLeS01

I get...
(x^2 - 4x + 4) - 4 + 20
(x - 2)^2 + 16

14. Jan 26, 2008

### rocomath

What does this Integral look like?

$$\int\frac{dx}{(x-2)^2+16}$$

If you don't plan on using an Integral table, factor out a 16 from the denominator b/c you need a constant of 1.

15. Jan 26, 2008

### BuBbLeS01

how do i factor out 16?? and why do I need a constant of 1?? I am so confused and I don't know why...I did all of the homework just fine.

16. Jan 26, 2008

### rocomath

B/c before you can integrate this integral, you need to have it in the form similar to ...

$$\int\frac{dx}{x^2+1}$$

Rather than factoring out a 16, divide both numerator & denominator by 16.

Last edited: Jan 26, 2008
17. Jan 26, 2008

### BuBbLeS01

Can anyone help me with this problem...I have a test Thursday and I am freaking out because I thought I understood this stuff!!

18. Jan 27, 2008

### rocomath

$$\int\frac{dx}{(x-2)^2+16}$$

To factor out a 16 ...

$$\frac{1}{16}\int\frac{dx}{\frac{(x-2)^2}{16}+1}$$

$$\frac{1}{16}\int\frac{dx}{\left(\frac{x-2}{4}\right)^2+1}$$

Last edited: Jan 27, 2008
19. Jan 27, 2008

### BuBbLeS01

Is there any other way to solve this problem? I am only asking because we have never done anything like that in class or on any homework and I find it hard to believe our teacher giving us something we haven't done before. So I was just curious if there was another solution?

20. Jan 27, 2008

### rocomath

It's no different from any of your other problems? You should be able to handle problems that are not 100% similar to what you have done b4. It's basically the integral of arctan.

Completing the square and factoring is something you learned prior to Calculus, so those 2 initial steps should not bother you.