What is the integral of (x+1)/(x^2-2x+5) ?

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

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integral of (x+1)/(x^2-2x+5)

\int \frac{x + 1}{x ^{2} - 2 x + 5} d x

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

Solution steps

\int \frac{x + 1}{x ^{2} - 2 x + 5} d x

Expand

\frac{x + 1}{x ^{2} - 2 x + 5} : \frac{x}{x ^{2} - 2 x + 5} + \frac{1}{x ^{2} - 2 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} - 2 x + 5} d x + \int \frac{1}{x ^{2} - 2 x + 5} d x

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

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

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

Add a constant to the solution

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

\int \frac{x ^{5} + x ^{4} - 8}{x ^{3} - x} d x

\int \frac{1}{x ( x ^{6} + 4 )} d x

\int \frac{x e ^{x}}{e ^{x} - 1} d x

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