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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x + 2} d x = ln ∣ x + 2 ∣

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} ((x + y) e^{y})

x \frac{d y}{d x} - 3 y = x^{4}

\int_{1}^{e} ln (5 x^{3}) d x

\int_{0}^{5} (7 x - x^{2}) - 2 x d x

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

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