How Do You Find the Anti-Derivative of (20/(1+x^2))^2?

[calculushelp]

I just don't get it.

how does

the anti derivative of

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

=

200arctan(x)+(200x)/(x^(2)+1)

 

[calculushelp]

oh p.s. that is how make TI-89 works it out.

 

[HallsofIvy]

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".

 

[swapnilster]

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.