- #1

- 13

- 0

how does

the anti derivative of

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

=

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

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- #1

- 13

- 0

how does

the anti derivative of

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

=

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

- #2

- 13

- 0

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

- #3

HallsofIvy

Science Advisor

Homework Helper

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It doesn't. That 200 is wrong.

how does

the anti derivative of

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

=

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

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

- 10

- 0

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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