# Forgot how to integrate fractions

1. ### Pengwuino

7,118
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. ### Benny

585
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. ### dextercioby

12,303
No need for any substitution. Benny's right.

Daniel.

4. ### Pengwuino

7,118
heh, i don't even remember the power rule.

5. ### Reshma

777
Power rule for integration is:
$$\int x^n dx = \frac{x^{n +1}}{n + 1} + C$$

6. ### dextercioby

12,303
Don't forget the condition

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

Daniel.

7. ### Nylex

551
Also remember that $$\frac{1}{x^n} = x^{-n}$$

8. ### disregardthat

1,811
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. ### arildno

12,015
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. ### Kruger

219
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. ### disregardthat

1,811
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. ### HallsofIvy

40,297
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. ### disregardthat

1,811
We all love parantheses!

14. ### emeraldevan

1
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

### 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.

### 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.