what is integral 1/(1-x)^2 dx, anyone could help?

Answer 1

Integral (1/(1-x)^2) dx

This one is actually fairly simple, because something linear is being squared. Note that 1/(1 - x)^2 is equal to (1 - x)^(-2).

Integral ( (1 - x)^(-2) )dx

We can use simple substitution to solve this.

Let u = 1 - x
du = (-1) dx
(-1) du = dx

And we substitute, to get

Integral (u^(-2) (-1)du)

Pulling out the constant -1, we get

(-1) * Integral (u^(-2))du

Which we can use the reverse power rule for.

(-1) [u^(-1) / (-1) ] + C

or

u^(-1) + C

or

1/u + C

and now we replace u = 1 - x, to get

1/(1 - x) + C

Answer 2

notes:y=1-x then dy=-dx and the integral is -1/y^2dy=1/y for y=1-x then we find the integral : 1/(1-x)

Answer 3

1-x=t dx=-dt
1/(1-x)^2dx=-1/t^2
==1/t==1/(1-x) +c

Answer 4

int 1/(1-x)^2 dx=int (1-x)^-2dx= 1/3*(1-x)^-3]+c