What is the integral of 8/((x+2)(x^2+4)) ?

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

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integral of 8/((x+2)(x^2+4))

\int \frac{8}{( x + 2 ) ( x ^{2} + 4 )} d x

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

Solution steps

\int \frac{8}{( x + 2 ) ( x ^{2} + 4 )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

\frac{1}{( x + 2 ) ( x ^{2} + 4 )} : \frac{1}{8 ( x + 2 )} + \frac{- x + 2}{8 ( x ^{2} + 4 )}

= 8 \cdot \int \frac{1}{8 ( x + 2 )} + \frac{- x + 2}{8 ( 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

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

\int \frac{1}{8 ( x + 2 )} d x = \frac{1}{8} ln ∣ x + 2 ∣

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

= 8 (\frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} (- \frac{1}{2} ln x^{2} + 4 + arctan (\frac{x}{2})))

Simplify

8 (\frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} (- \frac{1}{2} ln x^{2} + 4 + arctan (\frac{x}{2}))) : ln ∣ x + 2 ∣ - \frac{1}{2} ln x^{2} + 4 + arctan (\frac{x}{2})

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

Add a constant to the solution

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

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\int_{1}^{7} π (x - 1) d x

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