What is the integral of x(2x+5)^{20} ?

The integral of x(2x+5)^{20} is 1/1848 (21(2x+5)^{22}-110(2x+5)^{21})+C

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integral of x(2x+5)^{20}

\int x (2 x + 5)^{20} d x

\frac{1}{1848} (21 (2 x + 5)^{22} - 110 (2 x + 5)^{21}) + C

Solution steps

\int x (2 x + 5)^{20} d x

Apply u-substitution

: \int \frac{u ^{20} ( u - 5 )}{4} d u

= \int \frac{u ^{20} ( u - 5 )}{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^{20} (u - 5) d u

Expand

u^{20} (u - 5) : u^{21} - 5 u^{20}

= \frac{1}{4} \cdot \int u^{21} - 5 u^{20} 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^{21} d u - \int 5 u^{20} d u)

\int u^{21} d u = \frac{u ^{22}}{22}

\int 5 u^{20} d u = \frac{5 u ^{21}}{21}

= \frac{1}{4} (\frac{u ^{22}}{22} - \frac{5 u ^{21}}{21})

Substitute back

u = 2 x + 5

= \frac{1}{4} (\frac{( 2 x + 5 ) ^{22}}{22} - \frac{5 ( 2 x + 5 ) ^{21}}{21})

Simplify

\frac{1}{4} (\frac{( 2 x + 5 ) ^{22}}{22} - \frac{5 ( 2 x + 5 ) ^{21}}{21}) : \frac{1}{1848} (21 (2 x + 5)^{22} - 110 (2 x + 5)^{21})

= \frac{1}{1848} (21 (2 x + 5)^{22} - 110 (2 x + 5)^{21})

Add a constant to the solution

= \frac{1}{1848} (21 (2 x + 5)^{22} - 110 (2 x + 5)^{21}) + C

\frac{\partial}{\partial x} (7 e^{x} sin (y))

x \to 2 lim (7 x - 4)

x \to 2 lim (\frac{x ^{3} + 1}{x + 1})

x \to 9 lim (\frac{e ^{5 x} - 1}{x})

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

What is the integral of x(2x+5)^{20} ?
The integral of x(2x+5)^{20} is 1/1848 (21(2x+5)^{22}-110(2x+5)^{21})+C