What is the integral of 8/(x(x+2)^3) ?

The integral of 8/(x(x+2)^3) is 2/(x+2)+2/((x+2)^2)+ln(1/2 |x|)-ln| x/2+1|+C

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integral of 8/(x(x+2)^3)

\int \frac{8}{x ( x + 2 ) ^{3}} d x

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

Solution steps

\int \frac{8}{x ( x + 2 ) ^{3}} 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 ( x + 2 ) ^{3}} d x

x + 2 = 2 (\frac{x}{2} + 1)

= 8 \cdot \int \frac{1}{x ( 2 ( \frac{x}{2} + 1 ) ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{4 ( 2 u - 2 ) u ^{3}} d u

= 8 \cdot \int \frac{1}{4 ( 2 u - 2 ) u ^{3}} 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}{( 2 u - 2 ) u ^{3}} d u

Take the partial fraction of

\frac{1}{( 2 u - 2 ) u ^{3}} : \frac{1}{2 ( u - 1 )} - \frac{1}{2 u} - \frac{1}{2 u ^{2}} - \frac{1}{2 u ^{3}}

= 8 \cdot \frac{1}{4} \cdot \int \frac{1}{2 ( u - 1 )} - \frac{1}{2 u} - \frac{1}{2 u ^{2}} - \frac{1}{2 u ^{3}} 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{1}{2 u} d u - \int \frac{1}{2 u ^{2}} d u - \int \frac{1}{2 u ^{3}} d u)

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

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

\int \frac{1}{2 u ^{2}} d u = - \frac{1}{2 u}

\int \frac{1}{2 u ^{3}} d u = - \frac{1}{4 u ^{2}}

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

Substitute back

u = \frac{x}{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (x ln (2 x))

x \to 2 + lim (x^{2})

b^{x}

\int \frac{2 x}{2 x + 1} d x

x y^{2} + 1 + x^{2} \frac{d y}{d x} = 0

What is the integral of 8/(x(x+2)^3) ?
The integral of 8/(x(x+2)^3) is 2/(x+2)+2/((x+2)^2)+ln(1/2 |x|)-ln| x/2+1|+C