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

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

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

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

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

Decimal

3.24902 \dots

Solution steps

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

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

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

Simplify

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

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

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 t}{t} d t

Apply Integration By Parts

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

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

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

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

Simplify

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

= \frac{1}{2} [\frac{ln 6 t + 36 t + 1 + 6 t 36 t + 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/(4t)+9) ?
The integral from 0 to 1 of sqrt(1/(4t)+9) is (ln(6+sqrt(37))+6sqrt(37))/(12)