What is the integral of (x^2-2x+1)/(1-x) ?

The integral of (x^2-2x+1)/(1-x) is -(x^2)/2+x+C

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integral of (x^2-2x+1)/(1-x)

\int \frac{x ^{2} - 2 x + 1}{1 - x} d x

- \frac{x ^{2}}{2} + x + C

Solution steps

\int \frac{x ^{2} - 2 x + 1}{1 - x} d x

Simplify

\frac{x ^{2} - 2 x + 1}{1 - x} : - x + 1

= \int - x + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int x d x + \int 1 d x

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

= - \frac{x ^{2}}{2} + x

Add a constant to the solution

= - \frac{x ^{2}}{2} + x + C

\int (sin (x))^{7} d x

cos (x^{3})

\int \frac{π}{2} sin (\frac{π}{6} t) d t

\int e^{c o t (x)} csc^{2} (x) d x

What is the integral of (x^2-2x+1)/(1-x) ?
The integral of (x^2-2x+1)/(1-x) is -(x^2)/2+x+C