Anti derivative

  • #1

Main Question or Discussion Point

I just don't get it.

how does

the anti derivative of

( (20)/(1+x^(2)) )^(2)

=

200arctan(x)+(200x)/(x^(2)+1)
 

Answers and Replies

  • #2
oh p.s. that is how make TI-89 works it out.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,791
919
I just don't get it.

how does

the anti derivative of

( (20)/(1+x^(2)) )^(2)

=

200arctan(x)+(200x)/(x^(2)+1)
It doesn't. That 200 is wrong.

The derivative of arctan(x) is 1/(x^2+ 1). The derivative of x/(x^2+ 1), using the quotient rule, is [(1)(x^2+1)- (x)(2x)]/(x^2+ 1)^2= (1- x^2)/(x^2+1)^2

Their sum is (x^2+ 1)/(x^2+ 1)^2+ (1- x^2)/(x^2+1)^2= 2/(x^2+1)^2. Multiplying by 10, not 200, would give that "20".
 
  • #4
To get the antiderivative, you could substitute x=tan y and so dx=(sec y)^2 and thus simplifying
you would end up with (400/sec^2 y) dy
=>(400cos^2 y) dy
=>400(1+cos2y)dy
and integrate this and in the end substitute y=arctan x to get your answer.
 

Related Threads for: Anti derivative

  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
4
Views
2K
Replies
2
Views
2K
Replies
13
Views
16K
Replies
13
Views
5K
Replies
7
Views
2K
  • Last Post
Replies
5
Views
5K
Top