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

The integral of (5x+1)^{19} is 1/100 (5x+1)^{20}+C

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

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

\frac{1}{100} (5 x + 1)^{20} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{19}}{5} d u

= \int \frac{u ^{19}}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int u^{19} d u

Apply the Power Rule

: \frac{u ^{20}}{20}

= \frac{1}{5} \cdot \frac{u ^{20}}{20}

Substitute back

u = 5 x + 1

= \frac{1}{5} \cdot \frac{( 5 x + 1 ) ^{20}}{20}

Simplify

\frac{1}{5} \cdot \frac{( 5 x + 1 ) ^{20}}{20} : \frac{1}{100} (5 x + 1)^{20}

= \frac{1}{100} (5 x + 1)^{20}

Add a constant to the solution

= \frac{1}{100} (5 x + 1)^{20} + C

x \to 2 + lim (\frac{x}{2 - x})

y = e^{4 x}

y^{^{''}} + y^{^{'}} = \frac{1}{4} cos (3 t)

\frac{\partial}{\partial x} (e^{2 y - x})

\frac{x ^{2}}{x - 1}

What is the integral of (5x+1)^{19} ?
The integral of (5x+1)^{19} is 1/100 (5x+1)^{20}+C