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

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

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integral of (x^2)/((x-3)^3)

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

ln ∣ x - 3 ∣ - \frac{6}{x - 3} - \frac{9}{2 ( x - 3 ) ^{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{( u + 3 ) ^{2}}{u ^{3}} d u

= \int \frac{( u + 3 ) ^{2}}{u ^{3}} d u

Expand

\frac{( u + 3 ) ^{2}}{u ^{3}} : \frac{1}{u} + \frac{6}{u ^{2}} + \frac{9}{u ^{3}}

= \int \frac{1}{u} + \frac{6}{u ^{2}} + \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{6}{u ^{2}} d u + \int \frac{9}{u ^{3}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{6}{u ^{2}} d u = - \frac{6}{u}

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

= ln ∣ u ∣ - \frac{6}{u} - \frac{9}{2 u ^{2}}

Substitute back

u = x - 3

= ln ∣ x - 3 ∣ - \frac{6}{x - 3} - \frac{9}{2 ( x - 3 ) ^{2}}

Add a constant to the solution

= ln ∣ x - 3 ∣ - \frac{6}{x - 3} - \frac{9}{2 ( x - 3 ) ^{2}} + C

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

f (x) = x^{3} - 7 x^{2}, at x = 0

\int \frac{x ^{6}}{x ^{14} - 7} d x

\int 4 cos^{5} (x) sin^{4} (x) d x

x \to 8 - lim (7 x - 7)

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