# Forgot how to integrate fractions

by Pengwuino
Tags: forgot, fractions, integrate
 PF Gold P: 7,120 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?
 P: 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?
 Sci Advisor HW Helper P: 11,895 No need for any substitution. Benny's right. Daniel.
 PF Gold P: 7,120 Forgot how to integrate fractions heh, i don't even remember the power rule.
 P: 777 Power rule for integration is: $$\int x^n dx = \frac{x^{n +1}}{n + 1} + C$$
 Sci Advisor HW Helper P: 11,895 Don't forget the condition $$\int x^{n} \ dx=\frac{x^{n+1}}{n+1} +C , n\in{\mathbb{C}-\{-1\}}$$ Daniel.
 P: 552 Also remember that $$\frac{1}{x^n} = x^{-n}$$
 Sci Advisor P: 1,741 I'm sorry to bring this up, but how do you integrate a fraction? If you have 4x/3, how do you integrate it?
 Sci Advisor HW Helper PF Gold P: 12,016 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: $$\frac{4}{3}*\frac{x^{2}}{2}+C=\frac{4}{3*2}x^{2}+C=\frac{2}{3}x^{2}+C$$
 P: 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)
 Sci Advisor P: 1,741 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? $$\int{\frac{4x}{3}dx} = \frac{4}{3}\int{x*dx}$$
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PF Gold
P: 39,339
 Quote by Jarle 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? $$\int{\frac{4x}{3}dx} = \frac{4}{3}\int{x*dx}$$

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!
 Sci Advisor P: 1,741 We all love parantheses!
 P: 1 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
Mentor
P: 21,215
 Quote by 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
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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.
 Mentor P: 11,619 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.

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