What is the integral of (x^2-3x+2)/(x-1) ?

The integral of (x^2-3x+2)/(x-1) is 1/2 (x-1)^2-x+1+C

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integral of (x^2-3x+2)/(x-1)

\int \frac{x ^{2} - 3 x + 2}{x - 1} d x

\frac{1}{2} (x - 1)^{2} - x + 1 + C

Solution steps

\int \frac{x ^{2} - 3 x + 2}{x - 1} d x

Apply u-substitution

: \int u - 1 d u

= \int u - 1 d u

Apply the Sum Rule:

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

= \int u d u - \int 1 d u

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

\int 1 d u = u

= \frac{u ^{2}}{2} - u

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{2} (x - 1)^{2} - x + 1 + C

\int \frac{x ^{2} - 3 x - 7}{( 2 x + 3 ) ( x + 1 ) ^{2}} d x

\frac{d}{d x} (\frac{x ^{2}}{x ^{4} + 1})

\frac{d y}{d x} = 2 y - 1

x^{2} \frac{d y}{d x} + y^{2} = x y

\int cot^{2} (x) csc (x) d x

What is the integral of (x^2-3x+2)/(x-1) ?
The integral of (x^2-3x+2)/(x-1) is 1/2 (x-1)^2-x+1+C