What is the integral of (x-1)/(x^2-2x+4) ?

The integral of (x-1)/(x^2-2x+4) is 1/2 ln|x^2-2x+4|+C

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integral of (x-1)/(x^2-2x+4)

\int \frac{x - 1}{x ^{2} - 2 x + 4} d x

\frac{1}{2} ln x^{2} - 2 x + 4 + C

Solution steps

\int \frac{x - 1}{x ^{2} - 2 x + 4} d x

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} - 2 x + 4

= \frac{1}{2} ln x^{2} - 2 x + 4

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 2 x + 4 + C

\frac{d y}{d x} = \frac{9 x}{10 y}

\frac{\partial}{\partial x} (\frac{V ( x )}{x})

\int \frac{x}{( x - 2 ) ( x ^{2} - 4 x + 5 ) ^{2}} d x

x \to 2 lim (x^{5} - 8 - 2)

\frac{d}{d x} (x^{- 3} + x^{- 5} - 2 x^{- 6})

What is the integral of (x-1)/(x^2-2x+4) ?
The integral of (x-1)/(x^2-2x+4) is 1/2 ln|x^2-2x+4|+C