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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{2}{3}

Decimal

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

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

= \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}

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