What is the integral of (x^2)/((x-3)^5) ?

The integral of (x^2)/((x-3)^5) is -1/(2(x-3)^2)-2/((x-3)^3)-9/(4(x-3)^4)+C

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integral of (x^2)/((x-3)^5)

\int \frac{x ^{2}}{( x - 3 ) ^{5}} d x

- \frac{1}{2 ( x - 3 ) ^{2}} - \frac{2}{( x - 3 ) ^{3}} - \frac{9}{4 ( x - 3 ) ^{4}} + C

Solution steps

\int \frac{x ^{2}}{( x - 3 ) ^{5}} d x

Apply u-substitution

: \int \frac{( u + 3 ) ^{2}}{u ^{5}} d u

= \int \frac{( u + 3 ) ^{2}}{u ^{5}} d u

Expand

\frac{( u + 3 ) ^{2}}{u ^{5}} : \frac{1}{u ^{3}} + \frac{6}{u ^{4}} + \frac{9}{u ^{5}}

= \int \frac{1}{u ^{3}} + \frac{6}{u ^{4}} + \frac{9}{u ^{5}} 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 ^{3}} d u + \int \frac{6}{u ^{4}} d u + \int \frac{9}{u ^{5}} d u

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

\int \frac{6}{u ^{4}} d u = - \frac{2}{u ^{3}}

\int \frac{9}{u ^{5}} d u = - \frac{9}{4 u ^{4}}

= - \frac{1}{2 u ^{2}} - \frac{2}{u ^{3}} - \frac{9}{4 u ^{4}}

Substitute back

u = x - 3

= - \frac{1}{2 ( x - 3 ) ^{2}} - \frac{2}{( x - 3 ) ^{3}} - \frac{9}{4 ( x - 3 ) ^{4}}

Add a constant to the solution

= - \frac{1}{2 ( x - 3 ) ^{2}} - \frac{2}{( x - 3 ) ^{3}} - \frac{9}{4 ( x - 3 ) ^{4}} + C

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What is the integral of (x^2)/((x-3)^5) ?
The integral of (x^2)/((x-3)^5) is -1/(2(x-3)^2)-2/((x-3)^3)-9/(4(x-3)^4)+C