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|+1/2 ln|(x^2)/4+1|+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 \frac{x}{2} - 1 + \frac{1}{2} ln \frac{x ^{2}}{4} + 1 + 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

Apply Integral Substitution

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

Take the constant out:

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

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

Take the partial fraction of

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{2 ( u - 1 )} d u = \frac{1}{2} ln ∣ u - 1 ∣

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

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

Substitute back

u = \frac{x}{2}

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

Simplify

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

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

Add a constant to the solution

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

- 3 y^{^{'}} + 9 x^{2} = 0

\frac{\partial}{\partial x} (6 y^{2} x)

(x x)^{^{'}}

\frac{d y}{d t} + 7 y = e^{4 t}, y (0) = 10

\frac{\partial}{\partial x} (e^{- a x^{2}})

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|+1/2 ln|(x^2)/4+1|+arctan(x/2)+C