[sonic23]

Why is the integral of x^-2 = (1/-1)*x^-1

BUT

(1-x)^-2 = (1-x)^-1 i.e. not (1/-1)*(1-x)^-1

Thanks

[marcusmerehay]

For the second one, you have to take into account the minus sign in front of the x.

You can check this by differentiating the solution.

[sonic23]

(Original post by marcusmerehay)
For the second one, you have to take into account the minus sign in front of the x.

You can check this by differentiating the solution.

yeh exactly. you differentiate (1-x)^-1 and you get (x-1)^-2 not (1-x)^-2.

So then why is it when you differentiate (1-x)^-2 you get (1-x)^-1. Shouldnt it be MINUS (1-x)^-1.

[raheem94]

(Original post by sonic23)
Why is the integral of x^-2 = (1/-1)*x^-1

BUT

(1-x)^-2 = (1-x)^-1 i.e. not (1/-1)*(1-x)^-1

Thanks

You need to consider the coefficient of 'x', which in this case is -1, hence the -1 multiplies with the -1 obtained due to the power to give 1.

[raheem94]

(Original post by sonic23)
yeh exactly. you differentiate (1-x)^-1 and you get (x-1)^-2 not (1-x)^-2.

So then why is it when you differentiate (1-x)^-2 you get (1-x)^-1. Shouldnt it be MINUS (1-x)^-1.

Differentiating gives

[sonic23]

(Original post by raheem94)


You need to consider the coefficient of 'x', which in this case is -1, hence the -1 multiplies with the -1 obtained due to the power to give 1.

Thanks

[sonic23]

(Original post by raheem94)
Differentiating gives

I was using Wolfram.

And just realised (x-1)^-2 = (1-x)^-2

Thanks anyway

[ztibor]

(Original post by sonic23)
I was using Wolfram.

And just realised (x-1)^-2 = (1-x)^-2

Thanks anyway

THe left therm is reciprocal of and the other
Your question is why is
Because both quadratic term is positive and the bases of the powers differ only a minus sign, but
that is