Determine the indefinite integral of f(x)=1/(5x^2-10x+5)

To determine the indefinite integral, we'll factorize by
5 the denominator.

5x^2-10x+5 = 5(x^2 - 2x +
1)

We notice that the denominator is the result of
expanding the square: x^2-2x+1 = (x-1)^2

We'll re-write the
integral:

Int f(x)dx = (1/5)*Int
dx/(x-1)^2

We'll use the techinque of changing the
variable. For this reason we'll substitute x-1 by t.

x-1 =
t

We'll differentiate both
sides:

(x-1)' = 1*dx

t' =
dt

So, dx = dt

We'll re-write
the integral in t:

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

Int dt/t^2 = Int
[t^(-2)]*dt

Int [t^(-2)]*dt = t^(-2+1)/(-2+1) = t^(-1)/-1 =
-1/t

But t =
x-1

(1/5)*Int dx/(x-1)^2 = -1/5(x-1) +
C

or

(1/5)*Int
dx/(x-1)^2 = 1/5(1-x) + C