What is the integral of x(x-1)^8 ?

The integral of x(x-1)^8 is ((x-1)^{10}}{10}+\frac{(x-1)^9)/9+C

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integral of x(x-1)^8

\int x (x - 1)^{8} d x

\frac{( x - 1 ) ^{10}}{10} + \frac{( x - 1 ) ^{9}}{9} + C

Solution steps

\int x (x - 1)^{8} d x

Apply u-substitution

: \int (u + 1) u^{8} d u

= \int (u + 1) u^{8} d u

Expand

(u + 1) u^{8} : u^{9} + u^{8}

= \int u^{9} + u^{8} 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 u^{9} d u + \int u^{8} d u

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

\int u^{8} d u = \frac{u ^{9}}{9}

= \frac{u ^{10}}{10} + \frac{u ^{9}}{9}

Substitute back

u = x - 1

= \frac{( x - 1 ) ^{10}}{10} + \frac{( x - 1 ) ^{9}}{9}

Add a constant to the solution

= \frac{( x - 1 ) ^{10}}{10} + \frac{( x - 1 ) ^{9}}{9} + C

4 x (3 - 2 x)^{5}, at 2

\int (a + b x)^{n} d x

\int (sin (x)) d x

\int arctan (2 y) d y

y = ln (\frac{1 + e ^{x}}{1 - e ^{x}})

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