The Student Room Group

Integration query

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
(edited 11 years ago)
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.
Reply 2
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.
Reply 3
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


(ax+b)n dx=1a(ax+b)n+1n+1+C \displaystyle \int (ax+b)^n \ dx = \frac1a \frac{(ax+b)^{n+1}}{n+1} + C

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.
Reply 4
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 (1x)1 (1-x)^{-1} gives (1x)2 (1-x)^{-2}
Reply 5
Original post by raheem94
(ax+b)n dx=1a(ax+b)n+1n+1+C \displaystyle \int (ax+b)^n \ dx = \frac1a \frac{(ax+b)^{n+1}}{n+1} + C

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
Reply 6
Original post by raheem94
Differentiating (1x)1 (1-x)^{-1} gives (1x)2 (1-x)^{-2}


I was using Wolfram.

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

Thanks anyway
Reply 7
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 (x1))2(x-1))^2 and the other (1x)2(1-x)^2
Your question is why is (x1)2=(1x)2(x-1)^2=(1-x)^2
Because both quadratic term is positive and the bases of the powers differ only a minus sign, but (1)2=1(-1)^2=1
that is
(x1)2=[(1x)]2=(1)2(1x)2=(1x)2(x-1)^2=[-(1-x)]^2=(-1)^2\cdot(1-x)^2=(1-x)^2

Quick Reply

Latest