What is the integral of (x-3)/(-x^2+4x-5) ?

The integral of (x-3)/(-x^2+4x-5) is -1/2 ln|x^2-4x+5|+arctan(x-2)+C

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integral of (x-3)/(-x^2+4x-5)

\int \frac{x - 3}{- x ^{2} + 4 x - 5} d x

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

Solution steps

\int \frac{x - 3}{- x ^{2} + 4 x - 5} d x

Expand

\frac{x - 3}{- x ^{2} + 4 x - 5} : \frac{x}{- x ^{2} + 4 x - 5} - \frac{3}{- x ^{2} + 4 x - 5}

Apply the Sum Rule:

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

= \int \frac{x}{- x ^{2} + 4 x - 5} d x - \int \frac{3}{- x ^{2} + 4 x - 5} d x

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

\int \frac{3}{- x ^{2} + 4 x - 5} d x = - 3 arctan (x - 2)

= - \frac{1}{2} ln x^{2} - 4 x + 5 - 2 arctan (x - 2) - (- 3 arctan (x - 2))

Simplify

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

Add a constant to the solution

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

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

\frac{d}{d x} (4 x^{3} - 66 x^{2} + 216 x)

\frac{d}{d x} (cos^{2} (x^{3}) - ln (x^{2} - 1))

\frac{x ^{3}}{6} - 3 x^{2} + \frac{( 77 x )}{6} - 8, at x = 5

t^{3} ln (t)

What is the integral of (x-3)/(-x^2+4x-5) ?
The integral of (x-3)/(-x^2+4x-5) is -1/2 ln|x^2-4x+5|+arctan(x-2)+C