What is the integral of x(5x+1)^{19} ?

The integral of x(5x+1)^{19} is 1/10500 (20(5x+1)^{21}-21(5x+1)^{20})+C

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integral of x(5x+1)^{19}

\int x (5 x + 1)^{19} d x

\frac{1}{10500} (20 (5 x + 1)^{21} - 21 (5 x + 1)^{20}) + C

Solution steps

\int x (5 x + 1)^{19} d x

Apply u-substitution

: \int \frac{u ^{19} ( u - 1 )}{25} d u

= \int \frac{u ^{19} ( u - 1 )}{25} d u

Take the constant out:

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

= \frac{1}{25} \cdot \int u^{19} (u - 1) d u

Expand

u^{19} (u - 1) : u^{20} - u^{19}

= \frac{1}{25} \cdot \int u^{20} - u^{19} 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}{25} (\int u^{20} d u - \int u^{19} d u)

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

\int u^{19} d u = \frac{u ^{20}}{20}

= \frac{1}{25} (\frac{u ^{21}}{21} - \frac{u ^{20}}{20})

Substitute back

u = 5 x + 1

= \frac{1}{25} (\frac{( 5 x + 1 ) ^{21}}{21} - \frac{( 5 x + 1 ) ^{20}}{20})

Simplify

\frac{1}{25} (\frac{( 5 x + 1 ) ^{21}}{21} - \frac{( 5 x + 1 ) ^{20}}{20}) : \frac{1}{10500} (20 (5 x + 1)^{21} - 21 (5 x + 1)^{20})

= \frac{1}{10500} (20 (5 x + 1)^{21} - 21 (5 x + 1)^{20})

Add a constant to the solution

= \frac{1}{10500} (20 (5 x + 1)^{21} - 21 (5 x + 1)^{20}) + C

\int e^{- b t} d t

\int \frac{5}{9 + 5 x} d x

\int_{0}^{a} x^{2} e^{x} d x

\frac{\partial}{\partial x} (6 x^{3} y^{2})

x \to 1 lim ((5 x - 1)^{- 2})

What is the integral of x(5x+1)^{19} ?
The integral of x(5x+1)^{19} is 1/10500 (20(5x+1)^{21}-21(5x+1)^{20})+C