What is the integral of (x^2)/((x-2)^{10)} ?

The integral of (x^2)/((x-2)^{10)} is -1/(7(x-2)^7)-1/(2(x-2)^8)-4/(9(x-2)^9)+C

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integral of (x^2)/((x-2)^{10)}

\int \frac{x ^{2}}{( x - 2 ) ^{10}} d x

- \frac{1}{7 ( x - 2 ) ^{7}} - \frac{1}{2 ( x - 2 ) ^{8}} - \frac{4}{9 ( x - 2 ) ^{9}} + C

Solution steps

\int \frac{x ^{2}}{( x - 2 ) ^{10}} d x

Apply u-substitution

= \int \frac{( u + 2 ) ^{2}}{u ^{10}} d u

Expand

\frac{( u + 2 ) ^{2}}{u ^{10}} : \frac{1}{u ^{8}} + \frac{4}{u ^{9}} + \frac{4}{u ^{10}}

= \int \frac{1}{u ^{8}} + \frac{4}{u ^{9}} + \frac{4}{u ^{10}} 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 ^{8}} d u + \int \frac{4}{u ^{9}} d u + \int \frac{4}{u ^{10}} d u

\int \frac{1}{u ^{8}} d u = - \frac{1}{7 u ^{7}}

\int \frac{4}{u ^{9}} d u = - \frac{1}{2 u ^{8}}

\int \frac{4}{u ^{10}} d u = - \frac{4}{9 u ^{9}}

= - \frac{1}{7 u ^{7}} - \frac{1}{2 u ^{8}} - \frac{4}{9 u ^{9}}

Substitute back

u = x - 2

= - \frac{1}{7 ( x - 2 ) ^{7}} - \frac{1}{2 ( x - 2 ) ^{8}} - \frac{4}{9 ( x - 2 ) ^{9}}

Add a constant to the solution

= - \frac{1}{7 ( x - 2 ) ^{7}} - \frac{1}{2 ( x - 2 ) ^{8}} - \frac{4}{9 ( x - 2 ) ^{9}} + C

What is the integral of (x^2)/((x-2)^{10)} ?
The integral of (x^2)/((x-2)^{10)} is -1/(7(x-2)^7)-1/(2(x-2)^8)-4/(9(x-2)^9)+C