What is the integral from 0 to 2 of 1/(sqrt(4x+1)) ?

The integral from 0 to 2 of 1/(sqrt(4x+1)) is 1

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integral from 0 to 2 of 1/(sqrt(4x+1))

\int_{0}^{2} \frac{1}{4 x + 1} d x

1

Solution steps

\int_{0}^{2} \frac{1}{4 x + 1} d x

Apply u-substitution

: \int_{1}^{9} \frac{1}{4 u} d u

= \int_{1}^{9} \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{1}^{9} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{1}^{9}

= \frac{1}{4} [2 u^{\frac{1}{2}}]_{1}^{9}

Apply radical rule:

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

= \frac{1}{4} [2 u]_{1}^{9}

Compute the boundaries:

4

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

Simplify

= 1

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What is the integral from 0 to 2 of 1/(sqrt(4x+1)) ?
The integral from 0 to 2 of 1/(sqrt(4x+1)) is 1