What is the integral from 0 to 3 of 6/(sqrt(7x+5)) ?

The integral from 0 to 3 of 6/(sqrt(7x+5)) is (6(2sqrt(26)-2sqrt(5)))/7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 3 of 6/(sqrt(7x+5))

\int_{0}^{3} \frac{6}{7 x + 5} d x

\frac{6 ( 2 26 - 2 5 )}{7}

Decimal

4.90791 \dots

Solution steps

\int_{0}^{3} \frac{6}{7 x + 5} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{3} \frac{1}{7 x + 5} d x

Apply u-substitution

: \int_{5}^{26} \frac{1}{7 u} d u

= 6 \cdot \int_{5}^{26} \frac{1}{7 u} d u

Take the constant out:

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

= 6 \cdot \frac{1}{7} \cdot \int_{5}^{26} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{5}^{26}

= 6 \cdot \frac{1}{7} [2 u^{\frac{1}{2}}]_{5}^{26}

Simplify

6 \cdot \frac{1}{7} [2 u^{\frac{1}{2}}]_{5}^{26} : \frac{6}{7} [2 u]_{5}^{26}

= \frac{6}{7} [2 u]_{5}^{26}

Compute the boundaries:

226 - 25

= \frac{6}{7} (226 - 25)

Simplify

= \frac{6 ( 2 26 - 2 5 )}{7}

\int_{4}^{12} (4 - x) (x - 12) d x

\int_{0}^{\frac{π}{3}} (sec (x))^{2} d x

\int_{0}^{1} \frac{x}{( x + 1 ) ^{4}} d x

\int_{0}^{4} sin (\frac{nπ x}{4}) d x

\int_{0}^{3} 12 d x

What is the integral from 0 to 3 of 6/(sqrt(7x+5)) ?
The integral from 0 to 3 of 6/(sqrt(7x+5)) is (6(2sqrt(26)-2sqrt(5)))/7