What is the integral of (x^2-5x+6)/(x^2+4) ?

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

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integral of (x^2-5x+6)/(x^2+4)

\int \frac{x ^{2} - 5 x + 6}{x ^{2} + 4} d x

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

Solution steps

\int \frac{x ^{2} - 5 x + 6}{x ^{2} + 4} d x

\frac{x ^{2} - 5 x + 6}{x ^{2} + 4} = \frac{- 5 x + 2}{x ^{2} + 4} + 1

= \int \frac{- 5 x + 2}{x ^{2} + 4} + 1 d x

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{- 5 x + 2}{x ^{2} + 4} d x + \int 1 d x

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

\int 1 d x = x

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

Add a constant to the solution

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

\frac{\partial}{\partial x} ((\frac{y}{x})^{\frac{3}{4}})

\frac{\partial}{\partial y} (e^{x} cos (2 y))

y^{^{'}} + 4 y = 32 x, y (0) = 3

\frac{d y}{d x} = x y + 5 x + 4 y + 20

\frac{d}{d x} (7 e^{- 5 x})

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