What is the integral from 0 to (ln(6))/4 of ((5e^{4u}))/(sqrt(e^{4u)+3)} ?

The integral from 0 to (ln(6))/4 of ((5e^{4u}))/(sqrt(e^{4u)+3)} is 5/2

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integral from 0 to (ln(6))/4 of ((5e^{4u}))/(sqrt(e^{4u)+3)}

\int_{0}^{\frac{l n ( 6 )}{4}} \frac{( 5 e ^{4 u} )}{e ^{4 u} + 3} d u

\frac{5}{2}

Solution steps

\int_{0}^{\frac{l n ( 6 )}{4}} \frac{5 e ^{4 u}}{e ^{4 u} + 3} d u

Take the constant out:

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

= 5 \cdot \int_{0}^{\frac{l n ( 6 )}{4}} \frac{e ^{4 u}}{e ^{4 u} + 3} d u

Apply u-substitution

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

= 5 \cdot \int_{4}^{9} \frac{1}{4 v} d v

Take the constant out:

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

= 5 \cdot \frac{1}{4} \cdot \int_{4}^{9} \frac{1}{v} d v

Apply the Power Rule

: [2 v^{\frac{1}{2}}]_{4}^{9}

= 5 \cdot \frac{1}{4} [2 v^{\frac{1}{2}}]_{4}^{9}

Simplify

5 \cdot \frac{1}{4} [2 v^{\frac{1}{2}}]_{4}^{9} : \frac{5}{4} [2 v]_{4}^{9}

= \frac{5}{4} [2 v]_{4}^{9}

Compute the boundaries:

2

= \frac{5}{4} \cdot 2

Simplify

= \frac{5}{2}

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What is the integral from 0 to (ln(6))/4 of ((5e^{4u}))/(sqrt(e^{4u)+3)} ?
The integral from 0 to (ln(6))/4 of ((5e^{4u}))/(sqrt(e^{4u)+3)} is 5/2