What is the integral from 3 to 6 of 1/(sqrt(8x+1)) ?

The integral from 3 to 6 of 1/(sqrt(8x+1)) is 1/2

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integral from 3 to 6 of 1/(sqrt(8x+1))

\int_{3}^{6} \frac{1}{8 x + 1} d x

\frac{1}{2}

Decimal

0.5

Solution steps

\int_{3}^{6} \frac{1}{8 x + 1} d x

Apply u-substitution

= \int_{25}^{49} \frac{1}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{25}^{49} \frac{1}{u} d u

Apply the Power Rule

= \frac{1}{8} [2 u^{\frac{1}{2}}]_{25}^{49}

Apply radical rule:

= \frac{1}{8} [2 u]_{25}^{49}

Compute the boundaries:

4

= \frac{1}{8} \cdot 4

Simplify

= \frac{1}{2}

What is the integral from 3 to 6 of 1/(sqrt(8x+1)) ?
The integral from 3 to 6 of 1/(sqrt(8x+1)) is 1/2