What is the integral of (x^2-6x+8)/(x^2+2x+5) ?

The integral of (x^2-6x+8)/(x^2+2x+5) is x+11/2 arctan(1/2 (x+1))-4ln|x^2+2x+5|+C

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integral of (x^2-6x+8)/(x^2+2x+5)

\int \frac{x ^{2} - 6 x + 8}{x ^{2} + 2 x + 5} d x

x + \frac{11}{2} arctan (\frac{1}{2} (x + 1)) - 4 ln x^{2} + 2 x + 5 + C

Solution steps

\int \frac{x ^{2} - 6 x + 8}{x ^{2} + 2 x + 5} d x

Expand

\frac{x ^{2} - 6 x + 8}{x ^{2} + 2 x + 5} : \frac{x ^{2}}{x ^{2} + 2 x + 5} - \frac{6 x}{x ^{2} + 2 x + 5} + \frac{8}{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 ^{2}}{x ^{2} + 2 x + 5} d x - \int \frac{6 x}{x ^{2} + 2 x + 5} d x + \int \frac{8}{x ^{2} + 2 x + 5} d x

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

\int \frac{6 x}{x ^{2} + 2 x + 5} d x = 6 (\frac{1}{2} ln x^{2} + 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x + 1)))

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

= - ln x^{2} + 2 x + 5 - \frac{3}{2} arctan (\frac{1}{2} (x + 1)) + x - 6 (\frac{1}{2} ln x^{2} + 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x + 1))) + 4 arctan (\frac{1}{2} (x + 1))

Simplify

- ln x^{2} + 2 x + 5 - \frac{3}{2} arctan (\frac{1}{2} (x + 1)) + x - 6 (\frac{1}{2} ln x^{2} + 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x + 1))) + 4 arctan (\frac{1}{2} (x + 1)) : x + \frac{11}{2} arctan (\frac{1}{2} (x + 1)) - 4 ln x^{2} + 2 x + 5

= x + \frac{11}{2} arctan (\frac{1}{2} (x + 1)) - 4 ln x^{2} + 2 x + 5

Add a constant to the solution

= x + \frac{11}{2} arctan (\frac{1}{2} (x + 1)) - 4 ln x^{2} + 2 x + 5 + C

\frac{\partial}{\partial t} (sin^{2} (t))

\int 5 x + 1 d x

\frac{s}{( s ^{2} + 25 ) ^{2}}

\int 3 x (5 - x) d x

x \to 0 lim (\frac{x}{2 tan ( x )})

What is the integral of (x^2-6x+8)/(x^2+2x+5) ?
The integral of (x^2-6x+8)/(x^2+2x+5) is x+11/2 arctan(1/2 (x+1))-4ln|x^2+2x+5|+C