What is the integral of (x(x-8))/((x-4)^3) ?

The integral of (x(x-8))/((x-4)^3) is ln|x-4|+8/((x-4)^2)+C

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integral of (x(x-8))/((x-4)^3)

\int \frac{x ( x - 8 )}{( x - 4 ) ^{3}} d x

ln ∣ x - 4 ∣ + \frac{8}{( x - 4 ) ^{2}} + C

Solution steps

\int \frac{x ( x - 8 )}{( x - 4 ) ^{3}} d x

Complete the square

x (x - 8) : (x - 4)^{2} - 16

= \int \frac{( x - 4 ) ^{2} - 16}{( x - 4 ) ^{3}} d x

Apply u-substitution

= \int \frac{u ^{2} - 16}{u ^{3}} d u

Expand

\frac{u ^{2} - 16}{u ^{3}} : \frac{1}{u} - \frac{16}{u ^{3}}

= \int \frac{1}{u} - \frac{16}{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

= \int \frac{1}{u} d u - \int \frac{16}{u ^{3}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{16}{u ^{3}} d u = - \frac{8}{u ^{2}}

= ln ∣ u ∣ - (- \frac{8}{u ^{2}})

Substitute back

u = x - 4

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

Simplify

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

Add a constant to the solution

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

What is the integral of (x(x-8))/((x-4)^3) ?
The integral of (x(x-8))/((x-4)^3) is ln|x-4|+8/((x-4)^2)+C