# How to integrate -2x/ (1+x^2)

## Homework Statement

how do i integrate:
(-2x)/(1+x^2)^2

## 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

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morphism
Homework Helper
That substitution is a good idea, but it looks like you're making a mistake afterwards. Maybe if you post your working we'll be able to show you where you went wrong.

u=1+x^2
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?

HallsofIvy
Homework Helper
First the 2 in the denominator of dx= du/2x will cancel the 2 in the original integral: you should have "integral (-1/u^2)du.

(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 back to the 1+ x2? Don't undo your substitution until after you have integrated!

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

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

Hootenanny
Staff Emeritus
Gold Member
the integral is 1/u
Not quite,

$$\frac{d}{du}\left(\frac{1}{u}\right) = -\frac{1}{u^2} \neq \frac{1}{u^2}$$

Hint,

$$\int\frac{du}{u^2} = \int u^{-2} du$$

$$\int\frac{-2x}{(1+x^2)^2}dx=\int\frac{-2x}{(1+x^2)^2}\frac{d(1+x^2)}{2x}=\int\frac{-1}{(1+x^2)^2}d(1+x^2)=-\frac{(1+x^2)^{-2+1}}{-2+1}+C=\frac{1}{1+x^2}+C$$
where $$d(1+x^2)=2xdx$$
$$dx=\frac{d(1+x^2)}{2x}$$

no i dont have to integrate u^-2
i have to integrate -u^-2
which is 1/u ???

$$\int\frac{-2x}{(1+x^2)^2}dx=\int\frac{-2x}{(1+x^2)^2}\frac{d(1+x^2)}{2x}=\int\frac{-1}{(1+x^2)^2}d(1+x^2)=-\frac{(1+x^2)^{-2+1}}{-2+1}+C=\frac{1}{1+x^2}+C$$
where $$d(1+x^2)=2xdx$$
$$dx=\frac{d(1+x^2)}{2x}$$
yes i thought that was right. thank you

Hootenanny
Staff Emeritus
Ahh, my mistake, you are indeed correct. Sorry 