- #1

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## Homework Statement

how do i integrate:

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

## Homework Equations

## The Attempt at a Solution

I used the substitution u=1+x^2

then i got that the integral is

1/[2(1+x^2)^2 +C

but i'm not sure if that's correct

thank you

- Thread starter sara_87
- Start date

- #1

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how do i integrate:

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

I used the substitution u=1+x^2

then i got that the integral is

1/[2(1+x^2)^2 +C

but i'm not sure if that's correct

thank you

- #2

morphism

Science Advisor

Homework Helper

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- #3

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so dx=du/2x

so =integral(1/2u^2)du

=integral{1/[2(1+x^2)^2}

and now im stuck because i realised where i went wrong

so how do i continue from here?

- #4

HallsofIvy

Science Advisor

Homework Helper

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(Some people find it easier not to solve for dx: du= 2x dx and you can substitute directly for the "2xdx" in the original integral. And don't forget the "-".)

More important, why in the world did you go

Can you integrate [itex]\int (1/u^2) du[/itex]? (Hint: that's a power of u.)

- #5

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the integral is 1/u so thats 1/(1+x^2)

is that the answer?

is that the answer?

- #6

Hootenanny

Staff Emeritus

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Not quite,the integral is 1/u

[tex]\frac{d}{du}\left(\frac{1}{u}\right) = -\frac{1}{u^2} \neq \frac{1}{u^2}[/tex]

Hint,

[tex]\int\frac{du}{u^2} = \int u^{-2} du[/tex]

- #7

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where [tex]d(1+x^2)=2xdx[/tex]

[tex]dx=\frac{d(1+x^2)}{2x}[/tex]

- #8

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no i dont have to integrate u^-2

i have to integrate -u^-2

which is 1/u ???

i have to integrate -u^-2

which is 1/u ???

- #9

- 763

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yes i thought that was right. thank you

where [tex]d(1+x^2)=2xdx[/tex]

[tex]dx=\frac{d(1+x^2)}{2x}[/tex]

- #10

Hootenanny

Staff Emeritus

Science Advisor

Gold Member

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Ahh, my mistake, you are indeed correct. Sorryno i dont have to integrate u^-2

i have to integrate -u^-2

which is 1/u ???

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