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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

x \to \frac{π}{4} lim (tan^{2} (x))

\int 9 tan (x) d x

x \to - 1 lim (\frac{x + 2}{x + 1})

\frac{6}{( s - 2 ) ^{6}}

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