What is the integral of 7sqrt(7x-1) ?

The integral of 7sqrt(7x-1) is 2/3 (7x-1)^{3/2}+C

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integral of 7sqrt(7x-1)

\int 7 7 x - 1 d x

\frac{2}{3} (7 x - 1)^{\frac{3}{2}} + C

Solution steps

\int 7 7 x - 1 d x

Take the constant out:

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

= 7 \cdot \int 7 x - 1 d x

Apply u-substitution

: \int \frac{u}{7} d u

= 7 \cdot \int \frac{u}{7} d u

Take the constant out:

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

= 7 \cdot \frac{1}{7} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= 7 \cdot \frac{1}{7} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 7 x - 1

= 7 \cdot \frac{1}{7} \cdot \frac{2}{3} (7 x - 1)^{\frac{3}{2}}

Simplify

7 \cdot \frac{1}{7} \cdot \frac{2}{3} (7 x - 1)^{\frac{3}{2}} : \frac{2}{3} (7 x - 1)^{\frac{3}{2}}

= \frac{2}{3} (7 x - 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (7 x - 1)^{\frac{3}{2}} + C

\int_{\frac{1}{2}}^{1} (x^{- 3} - 13) d x

\frac{d}{d x} (4 sin (7 x))

x \to 0 lim (\frac{2 x + 1}{x + 3})

\frac{d y}{d x} = (y^{2} - 3 y)

\frac{\partial}{\partial x} (x^{\frac{1}{2}} y^{\frac{3}{2}})

What is the integral of 7sqrt(7x-1) ?
The integral of 7sqrt(7x-1) is 2/3 (7x-1)^{3/2}+C