What is the integral from 0 to 1 of sqrt(1/(4x)+9) ?

The integral from 0 to 1 of sqrt(1/(4x)+9) is (ln(6+sqrt(37))+6sqrt(37))/(12)

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integral from 0 to 1 of sqrt(1/(4x)+9)

\int_{0}^{1} \frac{1}{4 x} + 9 d x

\frac{ln ( 6 + 37 ) + 6 37}{12}

Decimal Notation

3.24902 \dots

Solution steps

\int_{0}^{1} \frac{1}{4 x} + 9 d x

\frac{1}{4 x} + 9 = 9 (\frac{\frac{1}{4 x}}{9} + 1)

= \int_{0}^{1} 9 (\frac{\frac{1}{4 x}}{9} + 1) d x

Simplify

9 (\frac{\frac{1}{4 x}}{9} + 1) : \frac{1 + 36 x}{2 x}

= \int_{0}^{1} \frac{1 + 36 x}{2 x} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{1} \frac{1 + 36 x}{x} d x

Apply Integration By Parts

: [2 x 1 + 36 x - \int \frac{36 x}{1 + 36 x} d x]_{0}^{1}

= \frac{1}{2} [2 x 1 + 36 x - \int \frac{36 x}{1 + 36 x} d x]_{0}^{1}

\int \frac{36 x}{1 + 36 x} d x = \frac{1}{6} (- ln 6 x + 36 x + 1 + 6 x 36 x + 1)

= \frac{1}{2} [2 x 1 + 36 x - \frac{1}{6} (- ln 6 x + 36 x + 1 + 6 x 36 x + 1)]_{0}^{1}

Simplify

\frac{1}{2} [2 x 1 + 36 x - \frac{1}{6} (- ln 6 x + 36 x + 1 + 6 x 36 x + 1)]_{0}^{1} : \frac{1}{2} [\frac{ln 6 x + 36 x + 1 + 6 x 36 x + 1}{6}]_{0}^{1}

= \frac{1}{2} [\frac{ln 6 x + 36 x + 1 + 6 x 36 x + 1}{6}]_{0}^{1}

Compute the boundaries:

\frac{ln ( 6 + 37 ) + 6 37}{6}

= \frac{1}{2} \cdot \frac{ln ( 6 + 37 ) + 6 37}{6}

Simplify

= \frac{ln ( 6 + 37 ) + 6 37}{12}

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What is the integral from 0 to 1 of sqrt(1/(4x)+9) ?
The integral from 0 to 1 of sqrt(1/(4x)+9) is (ln(6+sqrt(37))+6sqrt(37))/(12)