What is the integral from 0 to 4 of 1/(sqrt(6x+1)) ?

The integral from 0 to 4 of 1/(sqrt(6x+1)) is 4/3

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integral from 0 to 4 of 1/(sqrt(6x+1))

\int_{0}^{4} \frac{1}{6 x + 1} d x

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

\int_{0}^{4} \frac{1}{6 x + 1} d x

Apply u-substitution

: \int_{1}^{25} \frac{1}{6 u} d u

= \int_{1}^{25} \frac{1}{6 u} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{1}^{25} \frac{1}{u} d u

Apply the Power Rule

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

= \frac{1}{6} [2 u^{\frac{1}{2}}]_{1}^{25}

Apply radical rule:

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

= \frac{1}{6} [2 u]_{1}^{25}

Compute the boundaries:

8

= \frac{1}{6} \cdot 8

Simplify

= \frac{4}{3}

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What is the integral from 0 to 4 of 1/(sqrt(6x+1)) ?
The integral from 0 to 4 of 1/(sqrt(6x+1)) is 4/3