What is the integral of (x(x-6))/((x-3)^3) ?

The integral of (x(x-6))/((x-3)^3) is ln|x-3|+9/(2(x-3)^2)+C

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integral of (x(x-6))/((x-3)^3)

\int \frac{x ( x - 6 )}{( x - 3 ) ^{3}} d x

ln ∣ x - 3 ∣ + \frac{9}{2 ( x - 3 ) ^{2}} + C

Solution steps

\int \frac{x ( x - 6 )}{( x - 3 ) ^{3}} d x

Complete the square

x (x - 6) : (x - 3)^{2} - 9

= \int \frac{( x - 3 ) ^{2} - 9}{( x - 3 ) ^{3}} d x

Apply u-substitution

: \int \frac{u ^{2} - 9}{u ^{3}} d u

= \int \frac{u ^{2} - 9}{u ^{3}} d u

Expand

\frac{u ^{2} - 9}{u ^{3}} : \frac{1}{u} - \frac{9}{u ^{3}}

= \int \frac{1}{u} - \frac{9}{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{9}{u ^{3}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{9}{u ^{3}} d u = - \frac{9}{2 u ^{2}}

= ln ∣ u ∣ - (- \frac{9}{2 u ^{2}})

Substitute back

u = x - 3

= ln ∣ x - 3 ∣ - (- \frac{9}{2 ( x - 3 ) ^{2}})

Simplify

= ln ∣ x - 3 ∣ + \frac{9}{2 ( x - 3 ) ^{2}}

Add a constant to the solution

= ln ∣ x - 3 ∣ + \frac{9}{2 ( x - 3 ) ^{2}} + C

\int \frac{3}{9 - 4 x ^{2}} d x

y^{^{''}} - y^{^{'}} - 6 y = - 4 e^{2 t}, y (0) = 0, y^{^{'}} (0) = 0

f (x) = \frac{2 - x ^{2}}{2 + x ^{2}}

f (x)

(- 5, 5), (5, - 5)

What is the integral of (x(x-6))/((x-3)^3) ?
The integral of (x(x-6))/((x-3)^3) is ln|x-3|+9/(2(x-3)^2)+C