What is the integral of (4x-8)/(x^2-4x+5) ?

The integral of (4x-8)/(x^2-4x+5) is 2ln|x^2-4x+5|+C

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integral of (4x-8)/(x^2-4x+5)

\int \frac{4 x - 8}{x ^{2} - 4 x + 5} d x

2 ln x^{2} - 4 x + 5 + C

Solution steps

\int \frac{4 x - 8}{x ^{2} - 4 x + 5} d x

Apply Integration By Parts

: (4 x - 8) arctan (x - 2) - \int 4 arctan (x - 2) d x

= (4 x - 8) arctan (x - 2) - \int 4 arctan (x - 2) d x

\int 4 arctan (x - 2) d x = 4 (x arctan (x - 2) - \frac{1}{2} ln x^{2} - 4 x + 5 - 2 arctan (x - 2))

= (4 x - 8) arctan (x - 2) - 4 (x arctan (x - 2) - \frac{1}{2} ln x^{2} - 4 x + 5 - 2 arctan (x - 2))

Simplify

(4 x - 8) arctan (x - 2) - 4 (x arctan (x - 2) - \frac{1}{2} ln x^{2} - 4 x + 5 - 2 arctan (x - 2)) : 2 ln x^{2} - 4 x + 5

= 2 ln x^{2} - 4 x + 5

Add a constant to the solution

= 2 ln x^{2} - 4 x + 5 + C

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What is the integral of (4x-8)/(x^2-4x+5) ?
The integral of (4x-8)/(x^2-4x+5) is 2ln|x^2-4x+5|+C