What is the integral of 5(2x-1)^4 ?

The integral of 5(2x-1)^4 is 1/2 (2x-1)^5+C

integral of 5(2x-1)^4

\int 5 (2 x - 1)^{4} d x

\frac{1}{2} (2 x - 1)^{5} + C

Solution steps

\int 5 (2 x - 1)^{4} d x

Take the constant out:

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

= 5 \cdot \int (2 x - 1)^{4} d x

Apply u-substitution

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

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int u^{4} d u

Apply the Power Rule

= 5 \cdot \frac{1}{2} \cdot \frac{u ^{5}}{5}

Substitute back

u = 2 x - 1

= 5 \cdot \frac{1}{2} \cdot \frac{( 2 x - 1 ) ^{5}}{5}

Simplify

5 \cdot \frac{1}{2} \cdot \frac{( 2 x - 1 ) ^{5}}{5} : \frac{1}{2} (2 x - 1)^{5}

= \frac{1}{2} (2 x - 1)^{5}

Add a constant to the solution

= \frac{1}{2} (2 x - 1)^{5} + C

x \to 0 lim (3 arctan (e^{x}))

\int (a - x)^{2} d x

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

\int 5 x sec^{2} (x) d x

\frac{d}{d x} (\frac{x ^{2} - 2 x - 1}{( x - 1 ) ^{2}})