Integrate x^2/(1+4•x^2)? Partial fractions

  • Thread starter randoreds
  • Start date
  • #1
24
0
One last question


to Integrate x^2/(1+4•x^2). I would assume you would do long division but 4x^2 is bigger than x^2. so would you either pull out a 1/4 and it would be 1/4 ∫ x^2/(1/4+•x^2) dx or would the first term when doing long division be 1/4? or am I just totally wrong and you wouldn't do long division because the top is smaller than the bottom?
 

Answers and Replies

  • #2
6,054
391
Yes, you do a long division and you get 1/4 + remainder. What is the remainder?
 
  • #3
24
0
Yes, you do a long division and you get 1/4 + remainder. What is the remainder?

Ok, finished it but it was ugly. And I think I messed up my algebra somewhere.
your remainder ends up being -1/4 so you get∫ 1/4 +( -1/4/ (1+4x^2))so I decided to pull out the 4 to get the form x^2+a^2 so you get 1/4x + ∫-1/4 / (4((1/4)+x^2)) --> -1/16∫ 1/((1/4)+x^2) -->-1/16(1/(1/2)) times arctan(x/(1/2)) + C
So my final answer is

1/4 X - 1/8 arctan(2x) + C

If someone could check my math, I would be so thankful!
 
  • #4
6,054
391
Your result is correct.
 

Related Threads on Integrate x^2/(1+4•x^2)? Partial fractions

Replies
8
Views
5K
Replies
11
Views
3K
Replies
4
Views
3K
Replies
4
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
6
Views
15K
  • Last Post
Replies
15
Views
8K
Replies
3
Views
12K
Top