What is the integral from 3 to 7 of 1/(sqrt(6x+7)) ?

The integral from 3 to 7 of 1/(sqrt(6x+7)) is 2/3

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integral from 3 to 7 of 1/(sqrt(6x+7))

\int_{3}^{7} \frac{1}{6 x + 7} d x

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

\int_{3}^{7} \frac{1}{6 x + 7} d x

Apply u-substitution

: \int_{25}^{49} \frac{1}{6 u} d u

= \int_{25}^{49} \frac{1}{6 u} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{25}^{49} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{25}^{49}

= \frac{1}{6} [2 u^{\frac{1}{2}}]_{25}^{49}

Apply radical rule:

= \frac{1}{6} [2 u]_{25}^{49}

Compute the boundaries:

4

= \frac{1}{6} \cdot 4

Simplify

= \frac{2}{3}

\frac{\partial}{\partial u} (\frac{v}{u - v})

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

\frac{d}{d x} (\frac{3}{2 3 x + 1})

x \to 0 lim (4 cos (2 x))

y^{^{''}} - y^{^{'}} = 6 x e^{x}

What is the integral from 3 to 7 of 1/(sqrt(6x+7)) ?
The integral from 3 to 7 of 1/(sqrt(6x+7)) is 2/3