What is the integral from 1 to 4 of 1/(sqrt(4x+9)) ?

The integral from 1 to 4 of 1/(sqrt(4x+9)) is (5-sqrt(13))/2

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integral from 1 to 4 of 1/(sqrt(4x+9))

\int_{1}^{4} \frac{1}{4 x + 9} d x

\frac{5 - 13}{2}

Decimal Notation

0.69722 \dots

Solution steps

\int_{1}^{4} \frac{1}{4 x + 9} d x

Apply u-substitution

: \int_{13}^{25} \frac{1}{4 u} d u

= \int_{13}^{25} \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_{13}^{25} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{13}^{25}

= \frac{1}{4} [2 u^{\frac{1}{2}}]_{13}^{25}

Apply radical rule:

= \frac{1}{4} [2 u]_{13}^{25}

Compute the boundaries:

10 - 213

= \frac{1}{4} (10 - 213)

Simplify

= \frac{5 - 13}{2}

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What is the integral from 1 to 4 of 1/(sqrt(4x+9)) ?
The integral from 1 to 4 of 1/(sqrt(4x+9)) is (5-sqrt(13))/2