What is the integral of x(3+2x)^{10} ?

The integral of x(3+2x)^{10} is 1/528 (11(3+2x)^{12}-36(3+2x)^{11})+C

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integral of x(3+2x)^{10}

\int x (3 + 2 x)^{10} d x

\frac{1}{528} (11 (3 + 2 x)^{12} - 36 (3 + 2 x)^{11}) + C

Solution steps

\int x (3 + 2 x)^{10} d x

Apply u-substitution

= \int \frac{u ^{10} ( u - 3 )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u^{10} (u - 3) d u

Expand

u^{10} (u - 3) : u^{11} - 3 u^{10}

= \frac{1}{4} \cdot \int u^{11} - 3 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

= \frac{1}{4} (\int u^{11} d u - \int 3 u^{10} d u)

\int u^{11} d u = \frac{u ^{12}}{12}

\int 3 u^{10} d u = \frac{3 u ^{11}}{11}

= \frac{1}{4} (\frac{u ^{12}}{12} - \frac{3 u ^{11}}{11})

Substitute back

u = 3 + 2 x

= \frac{1}{4} (\frac{( 3 + 2 x ) ^{12}}{12} - \frac{3 ( 3 + 2 x ) ^{11}}{11})

Simplify

\frac{1}{4} (\frac{( 3 + 2 x ) ^{12}}{12} - \frac{3 ( 3 + 2 x ) ^{11}}{11}) : \frac{1}{528} (11 (3 + 2 x)^{12} - 36 (3 + 2 x)^{11})

= \frac{1}{528} (11 (3 + 2 x)^{12} - 36 (3 + 2 x)^{11})

Add a constant to the solution

= \frac{1}{528} (11 (3 + 2 x)^{12} - 36 (3 + 2 x)^{11}) + C

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What is the integral of x(3+2x)^{10} ?
The integral of x(3+2x)^{10} is 1/528 (11(3+2x)^{12}-36(3+2x)^{11})+C