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Forgot how to integrate fractions 
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#1
Nov305, 04:03 AM

PF Gold
P: 7,120

Ok so I have to find the arc length from t=1 to t=2.
[tex] \begin{array}{l} L = \int_a^b {r'(t)dt} \\ r'(t) = \frac{{2(1 + 2t^4 )}}{{t^3 }} \\ \end{array} [/tex] 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? 


#2
Nov305, 04:07 AM

P: 585

I hope I'm not missing something.
[tex]\frac{{2\left( {1 + 2t^4 } \right)}}{{t^3 }} = 2\left( {\frac{1}{{t^3 }} + \frac{{2t^4 }}{{t^3 }}} \right)[/tex] then it should just be the normal power rule right? 


#3
Nov305, 04:16 AM

Sci Advisor
HW Helper
P: 11,915

No need for any substitution. Benny's right.
Daniel. 


#4
Nov305, 04:20 AM

PF Gold
P: 7,120

Forgot how to integrate fractions
heh, i don't even remember the power rule.



#5
Nov305, 04:33 AM

P: 777

Power rule for integration is:
[tex]\int x^n dx = \frac{x^{n +1}}{n + 1} + C[/tex] 


#6
Nov305, 04:43 AM

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P: 11,915

Don't forget the condition
[tex] \int x^{n} \ dx=\frac{x^{n+1}}{n+1} +C , n\in{\mathbb{C}\{1\}} [/tex] Daniel. 


#7
Nov305, 05:31 AM

P: 551

Also remember that [tex]\frac{1}{x^n} = x^{n}[/tex]



#8
Mar607, 03:35 PM

Sci Advisor
P: 1,797

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
Mar607, 03:40 PM

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PF Gold
P: 12,016

Remember that a fraction is nothing else than an ordinary number. Also remember that you can write:
[tex]\frac{4x}{3}=\frac{4}{3}*x[/tex] f Thus, an antiderivative is: [tex]\frac{4}{3}*\frac{x^{2}}{2}+C=\frac{4}{3*2}x^{2}+C=\frac{2}{3}x^{2}+C[/tex] 


#10
Mar607, 05:00 PM

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) 


#11
Mar707, 11:00 AM

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P: 1,797

Thanks,
[tex]\int{C*f(x)dx} = C[/tex] [tex]\int{f(x)dx}[/tex] But Kruger, wouldn't the dx be over/beside the fraction, and not under: like this? [tex]\int{\frac{4x}{3}dx} = \frac{4}{3}\int{x*dx}[/tex] 


#12
Mar707, 01:35 PM

Math
Emeritus
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Thanks
PF Gold
P: 39,510

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
Mar807, 08:40 AM

Sci Advisor
P: 1,797

We all love parantheses!



#14
Mar2311, 05:05 PM

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 xaxis The curve eqn is 6/(52x) and the x limits are 0 and 1 


#15
Mar2311, 05:14 PM

Mentor
P: 21,265

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. 


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