# Forgot how to integrate fractions

1. Nov 3, 2005

### Pengwuino

Ok so I have to find the arc length from t=1 to t=2.

$$\begin{array}{l} L = \int_a^b {|r'(t)|dt} \\ |r'(t)| = \frac{{2(1 + 2t^4 )}}{{t^3 }} \\ \end{array}$$

And I have completely forgotten how to integrate fractions.....

Oh wait... i THINK i know what to do. Should i set u=1+2t^4?

Last edited: Nov 3, 2005
2. Nov 3, 2005

### Benny

I hope I'm not missing something.

$$\frac{{2\left( {1 + 2t^4 } \right)}}{{t^3 }} = 2\left( {\frac{1}{{t^3 }} + \frac{{2t^4 }}{{t^3 }}} \right)$$ then it should just be the normal power rule right?

3. Nov 3, 2005

### dextercioby

No need for any substitution. Benny's right.

Daniel.

4. Nov 3, 2005

### Pengwuino

heh, i don't even remember the power rule.

5. Nov 3, 2005

### Reshma

Power rule for integration is:
$$\int x^n dx = \frac{x^{n +1}}{n + 1} + C$$

6. Nov 3, 2005

### dextercioby

Don't forget the condition

$$\int x^{n} \ dx=\frac{x^{n+1}}{n+1} +C , n\in{\mathbb{C}-\{-1\}}$$

Daniel.

7. Nov 3, 2005

### Nylex

Also remember that $$\frac{1}{x^n} = x^{-n}$$

8. Mar 6, 2007

### disregardthat

I'm sorry to bring this up, but how do you integrate a fraction?

If you have 4x/3, how do you integrate it?

9. Mar 6, 2007

### arildno

Remember that a fraction is nothing else than an ordinary number. Also remember that you can write:
$$\frac{4x}{3}=\frac{4}{3}*x$$
f
Thus, an anti-derivative is:
$$\frac{4}{3}*\frac{x^{2}}{2}+C=\frac{4}{3*2}x^{2}+C=\frac{2}{3}x^{2}+C$$

10. Mar 6, 2007

### Kruger

Also think of the following: If there is a constant inside an integral you can move it outside the integral (sorry, don't know this formula editor):

INTEGRAL(c*f(x)dx)=c*INTEGRAL(f(x)dx)

So in your case 4/3 is a constant (independant of x) and you can move it outside the integral, getting:

INTEGRAL(4x/3dx)=4/3*INTEGRAL(xdx)

11. Mar 7, 2007

### disregardthat

Thanks,

$$\int{C*f(x)dx} = C$$ $$\int{f(x)dx}$$

But Kruger, wouldn't the dx be over/beside the fraction, and not under: like this?

$$\int{\frac{4x}{3}dx} = \frac{4}{3}\int{x*dx}$$

Last edited: Mar 7, 2007
12. Mar 7, 2007

### HallsofIvy

Staff Emeritus

That's what he meant. Most people interpret 4/3a to be (4/3)a, not 4/(3a).
It's always best to use the parentheses, though!

13. Mar 8, 2007

### disregardthat

We all love parantheses!

14. Mar 23, 2011

### emeraldevan

Hi, found this thread and hoping to get a reply:
I want to integrate curve eqn to get volume of curve rotated around x-axis
The curve eqn is 6/(5-2x) and the x limits are 0 and 1

15. Mar 23, 2011

### Staff: Mentor

Welcome to Physics Forums!

This thread is more than four years old. When you tack an unrelated question onto an existing thread, that's called "hijacking" the thread. Please use the New Thread button to start a new thread.

16. Mar 23, 2011

### Staff: Mentor

Actually, the button is labeled "New Topic" not "New Thread", otherwise Mark's advice is good. Also, please show the work that you have already done on the problem, or at least tell us what is confusing you about how to solve it.