What is the integral from 0 to 1 of 1/(sqrt(3x+1)) ?

The integral from 0 to 1 of 1/(sqrt(3x+1)) is 2/3

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integral from 0 to 1 of 1/(sqrt(3x+1))

\int_{0}^{1} \frac{1}{3 x + 1} d x

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

\int_{0}^{1} \frac{1}{3 x + 1} d x

Apply u-substitution

: \int_{1}^{4} \frac{1}{3 u} d u

= \int_{1}^{4} \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_{1}^{4} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{1}^{4}

= \frac{1}{3} [2 u^{\frac{1}{2}}]_{1}^{4}

Apply radical rule:

a^{\frac{1}{n}} = n a

= \frac{1}{3} [2 u]_{1}^{4}

Compute the boundaries:

2

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

Simplify

= \frac{2}{3}

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What is the integral from 0 to 1 of 1/(sqrt(3x+1)) ?
The integral from 0 to 1 of 1/(sqrt(3x+1)) is 2/3