What is the integral from 0 to 5 of 2/(sqrt(3x+2)) ?

The integral from 0 to 5 of 2/(sqrt(3x+2)) is (2(2sqrt(17)-2sqrt(2)))/3

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integral from 0 to 5 of 2/(sqrt(3x+2))

\int_{0}^{5} \frac{2}{3 x + 2} d x

\frac{2 ( 2 17 - 2 2 )}{3}

Decimal

3.61185 \dots

Solution steps

\int_{0}^{5} \frac{2}{3 x + 2} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{5} \frac{1}{3 x + 2} d x

Apply u-substitution

= 2 \cdot \int_{2}^{17} \frac{1}{3 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int_{2}^{17} \frac{1}{u} d u

Apply the Power Rule

= 2 \cdot \frac{1}{3} [2 u^{\frac{1}{2}}]_{2}^{17}

Simplify

2 \cdot \frac{1}{3} [2 u^{\frac{1}{2}}]_{2}^{17} : \frac{2}{3} [2 u]_{2}^{17}

= \frac{2}{3} [2 u]_{2}^{17}

Compute the boundaries:

217 - 22

= \frac{2}{3} (217 - 22)

Simplify

= \frac{2 ( 2 17 - 2 2 )}{3}

What is the integral from 0 to 5 of 2/(sqrt(3x+2)) ?
The integral from 0 to 5 of 2/(sqrt(3x+2)) is (2(2sqrt(17)-2sqrt(2)))/3