What is the integral of (e^{12t})/(sqrt(4e^{6t)-1)} ?

The integral of (e^{12t})/(sqrt(4e^{6t)-1)} is 1/72 (2e^{6t}+1)sqrt(4e^{6t)-1}+C

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integral of (e^{12t})/(sqrt(4e^{6t)-1)}

\int \frac{e ^{12 t}}{4 e ^{6 t} - 1} d t

\frac{1}{72} (2 e^{6 t} + 1) 4 e^{6 t} - 1 + C

Solution steps

\int \frac{e ^{12 t}}{4 e ^{6 t} - 1} d t

Apply u-substitution

: \int \frac{u + 1}{96 u} d u

= \int \frac{u + 1}{96 u} d u

Take the constant out:

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

= \frac{1}{96} \cdot \int \frac{u + 1}{u} d u

Expand

\frac{u + 1}{u} : u + \frac{1}{u}

= \frac{1}{96} \cdot \int u + \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{96} (\int u d u + \int \frac{1}{u} d u)

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

\int \frac{1}{u} d u = 2 u

= \frac{1}{96} (\frac{2}{3} u^{\frac{3}{2}} + 2 u)

Substitute back

u = 4 e^{6 t} - 1

= \frac{1}{96} (\frac{2}{3} (4 e^{6 t} - 1)^{\frac{3}{2}} + 2 4 e^{6 t} - 1)

Simplify

\frac{1}{96} (\frac{2}{3} (4 e^{6 t} - 1)^{\frac{3}{2}} + 2 4 e^{6 t} - 1) : \frac{1}{72} (2 e^{6 t} + 1) 4 e^{6 t} - 1

= \frac{1}{72} (2 e^{6 t} + 1) 4 e^{6 t} - 1

Add a constant to the solution

= \frac{1}{72} (2 e^{6 t} + 1) 4 e^{6 t} - 1 + C

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What is the integral of (e^{12t})/(sqrt(4e^{6t)-1)} ?
The integral of (e^{12t})/(sqrt(4e^{6t)-1)} is 1/72 (2e^{6t}+1)sqrt(4e^{6t)-1}+C