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

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### 2 Answers

To find the indefinite integral of(x) = 1/5x^2-10x+5.

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

f(x) = 1/5(x^2-2x+1)

f(x) =(1/5)(x-1)^-2.

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

Int (1/5)(x-1)^-2 dx = (1/5)int t^-2 dt.....(1) where (x-1)=t dx = dt

Now we use Int t^n dt = (1/(n+1))t^(n+1).

Therefore Int t^-2 dt = (1/(-2+1)) t^(-2+1).

So we rewrite (1) : (1/5) int (x-1)^-2 dx = (1/5) int t^(-2) dt = (1/5(-1))( t^-1) +C

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

Int 1/(5x^2-10x+5) dx = -1/(5(x-1)) +C

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**