What is the integral of (x^2)/((1-x)^{100)} ?

The integral of (x^2)/((1-x)^{100)} is 1/(97(1-x)^{97)}-1/(49(1-x)^{98)}+1/(99(1-x)^{99)}+C

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integral of (x^2)/((1-x)^{100)}

\int \frac{x ^{2}}{( 1 - x ) ^{100}} d x

\frac{1}{97 ( 1 - x ) ^{97}} - \frac{1}{49 ( 1 - x ) ^{98}} + \frac{1}{99 ( 1 - x ) ^{99}} + C

Solution steps

\int \frac{x ^{2}}{( 1 - x ) ^{100}} d x

Apply u-substitution

: \int - \frac{( - u + 1 ) ^{2}}{u ^{100}} d u

= \int - \frac{( - u + 1 ) ^{2}}{u ^{100}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{( - u + 1 ) ^{2}}{u ^{100}} d u

Expand

\frac{( - u + 1 ) ^{2}}{u ^{100}} : \frac{1}{u ^{98}} - \frac{2}{u ^{99}} + \frac{1}{u ^{100}}

= - \int \frac{1}{u ^{98}} - \frac{2}{u ^{99}} + \frac{1}{u ^{100}} 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 ^{98}} d u - \int \frac{2}{u ^{99}} d u + \int \frac{1}{u ^{100}} d u)

\int \frac{1}{u ^{98}} d u = - \frac{1}{97 u ^{97}}

\int \frac{2}{u ^{99}} d u = - \frac{1}{49 u ^{98}}

\int \frac{1}{u ^{100}} d u = - \frac{1}{99 u ^{99}}

= - (- \frac{1}{97 u ^{97}} - (- \frac{1}{49 u ^{98}}) - \frac{1}{99 u ^{99}})

Substitute back

u = 1 - x

= - (- \frac{1}{97 ( 1 - x ) ^{97}} - (- \frac{1}{49 ( 1 - x ) ^{98}}) - \frac{1}{99 ( 1 - x ) ^{99}})

Simplify

= \frac{1}{97 ( 1 - x ) ^{97}} - \frac{1}{49 ( 1 - x ) ^{98}} + \frac{1}{99 ( 1 - x ) ^{99}}

Add a constant to the solution

= \frac{1}{97 ( 1 - x ) ^{97}} - \frac{1}{49 ( 1 - x ) ^{98}} + \frac{1}{99 ( 1 - x ) ^{99}} + C

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What is the integral of (x^2)/((1-x)^{100)} ?
The integral of (x^2)/((1-x)^{100)} is 1/(97(1-x)^{97)}-1/(49(1-x)^{98)}+1/(99(1-x)^{99)}+C