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

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

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

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

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

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

Apply u-substitution

: \int_{16}^{25} \frac{1}{3 u} d u

= \int_{16}^{25} \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{16}^{25} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{16}^{25}

= \frac{1}{3} [2 u^{\frac{1}{2}}]_{16}^{25}

Apply radical rule:

= \frac{1}{3} [2 u]_{16}^{25}

Compute the boundaries:

2

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

Simplify

= \frac{2}{3}

\frac{d}{d x} ((3 x^{4}) (4 x^{3} + 4 x))

\int \frac{e ^{2 x}}{e ^{x} - 5} d x

f (x) = (7 x^{2} - 12 x) e^{x}

3 x^{2}, x^{2} + 6

\frac{d y}{d x} = 6 x y

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