What is the integral of x(x-4)^9 ?

The integral of x(x-4)^9 is ((x-4)^{11}}{11}+\frac{2(x-4)^{10})/5+C

integral of x(x-4)^9

\int x (x - 4)^{9} d x

\frac{( x - 4 ) ^{11}}{11} + \frac{2 ( x - 4 ) ^{10}}{5} + C

Solution steps

\int x (x - 4)^{9} d x

Apply u-substitution

= \int (u + 4) u^{9} d u

Expand

(u + 4) u^{9} : u^{10} + 4 u^{9}

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

\int u^{10} d u = \frac{u ^{11}}{11}

\int 4 u^{9} d u = \frac{2 u ^{10}}{5}

= \frac{u ^{11}}{11} + \frac{2 u ^{10}}{5}

Substitute back

u = x - 4

= \frac{( x - 4 ) ^{11}}{11} + \frac{2 ( x - 4 ) ^{10}}{5}

Add a constant to the solution

= \frac{( x - 4 ) ^{11}}{11} + \frac{2 ( x - 4 ) ^{10}}{5} + C

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What is the integral of x(x-4)^9 ?
The integral of x(x-4)^9 is ((x-4)^{11}}{11}+\frac{2(x-4)^{10})/5+C