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

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SUMMARY

The anti-derivative of the function (20/(1+x^2))^2 is incorrectly stated as 200arctan(x)+(200x)/(x^2+1). The correct approach involves recognizing that the derivative of arctan(x) is 1/(x^2+1) and applying the quotient rule to find the derivative of x/(x^2+1). The accurate anti-derivative can be derived by substituting x=tan(y) and simplifying the integral, ultimately leading to the correct result.

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I just don't get it.

how does

the anti derivative of

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

=

200arctan(x)+(200x)/(x^(2)+1)
 
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oh p.s. that is how make TI-89 works it out.
 
calculushelp said:
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".
 
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.
 

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