What is the integral of (x^2-x+10)/((x+2)(x^2+4)) ?

The integral of (x^2-x+10)/((x+2)(x^2+4)) is 2ln|x+2|-1/2 ln|x^2+4|+1/2 arctan(x/2)+C

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integral of (x^2-x+10)/((x+2)(x^2+4))

\int \frac{x ^{2} - x + 10}{( x + 2 ) ( x ^{2} + 4 )} d x

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

Solution steps

\int \frac{x ^{2} - x + 10}{( x + 2 ) ( x ^{2} + 4 )} d x

Take the partial fraction of

\frac{x ^{2} - x + 10}{( x + 2 ) ( x ^{2} + 4 )} : \frac{2}{x + 2} + \frac{- x + 1}{x ^{2} + 4}

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

\int \frac{2}{x + 2} d x = 2 ln ∣ x + 2 ∣

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (2 x^{3} + x^{2} y^{3} - 3 y^{2})

\frac{d y}{d x} = \frac{cos ( 14 x )}{e ^{14 y}}

n = 0 \sum \infty 3 (- 0.1)^{n}

\int arctan (u) d u

\int_{- 4}^{4} 4 (16 - x^{2}) d x

What is the integral of (x^2-x+10)/((x+2)(x^2+4)) ?
The integral of (x^2-x+10)/((x+2)(x^2+4)) is 2ln|x+2|-1/2 ln|x^2+4|+1/2 arctan(x/2)+C