What is the integral of (x^2)/((x+1)^{10)} ?

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

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integral of (x^2)/((x+1)^{10)}

\int \frac{x ^{2}}{( x + 1 ) ^{10}} d x

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

Solution steps

\int \frac{x ^{2}}{( x + 1 ) ^{10}} d x

Apply u-substitution

: \int \frac{( u - 1 ) ^{2}}{u ^{10}} d u

= \int \frac{( u - 1 ) ^{2}}{u ^{10}} d u

Expand

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

= \int \frac{1}{u ^{8}} - \frac{2}{u ^{9}} + \frac{1}{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{2}{u ^{9}} d u + \int \frac{1}{u ^{10}} d u

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

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

\int \frac{1}{u ^{10}} d u = - \frac{1}{9 u ^{9}}

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

Substitute back

u = x + 1

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

Simplify

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

Add a constant to the solution

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

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