What is the integral of 2sqrt(2x+7) ?

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

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integral of 2sqrt(2x+7)

\int 2 2 x + 7 d x

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

Solution steps

\int 2 2 x + 7 d x

Take the constant out:

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

= 2 \cdot \int 2 x + 7 d x

Apply u-substitution

: \int \frac{u}{2} d u

= 2 \cdot \int \frac{u}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int u d u

Apply the Power Rule

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

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

Substitute back

u = 2 x + 7

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

Simplify

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

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

Add a constant to the solution

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

\int (x x) d x

f (x) = - 24 π sin (π x)

y = x^{2} + 4 x^{3}

\frac{1}{y ( 1 - 0.0005 y )} d y = d x, y (0) = 1

\frac{d}{d x} (\frac{5}{cos ( x )})

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