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

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

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

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

- \frac{1}{x - 1} + C

Solution steps

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

Complete the square

1 - 2 x + x^{2} : (x - 1)^{2}

= \int \frac{1}{( x - 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

= \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = x - 1

= - \frac{1}{x - 1}

Add a constant to the solution

= - \frac{1}{x - 1} + C

\int e^{- 2 x} sin (4 x) d x

6 (t - 4)

\frac{\partial}{\partial x} (e^{\frac{1}{x}})

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

\frac{d y}{d t} + 3 y = 11, y (0) = 1

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