What is the integral of 1/(sqrt(4^x-4)) ?

The integral of 1/(sqrt(4^x-4)) is 1/(2ln(2))arctan(1/2 sqrt(4^x-4))+C

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integral of 1/(sqrt(4^x-4))

\int \frac{1}{4 ^{x} - 4} d x

\frac{1}{2 ln ( 2 )} arctan (\frac{1}{2} 4^{x} - 4) + C

Solution steps

\int \frac{1}{4 ^{x} - 4} d x

Apply u-substitution

: \int \frac{1}{ln ( 2 ) ( u ^{2} + 4 )} d u

= \int \frac{1}{ln ( 2 ) ( u ^{2} + 4 )} d u

Take the constant out:

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

= \frac{1}{ln ( 2 )} \cdot \int \frac{1}{u ^{2} + 4} d u

Apply Integral Substitution

= \frac{1}{ln ( 2 )} \cdot \int \frac{1}{2 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{ln ( 2 )} \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{ln ( 2 )} \cdot \frac{1}{2} arctan (v)

Substitute back

= \frac{1}{ln ( 2 )} \cdot \frac{1}{2} arctan (\frac{4 ^{x} - 4}{2})

Simplify

\frac{1}{ln ( 2 )} \cdot \frac{1}{2} arctan (\frac{4 ^{x} - 4}{2}) : \frac{1}{2 ln ( 2 )} arctan (\frac{1}{2} 4^{x} - 4)

= \frac{1}{2 ln ( 2 )} arctan (\frac{1}{2} 4^{x} - 4)

Add a constant to the solution

= \frac{1}{2 ln ( 2 )} arctan (\frac{1}{2} 4^{x} - 4) + C

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What is the integral of 1/(sqrt(4^x-4)) ?
The integral of 1/(sqrt(4^x-4)) is 1/(2ln(2))arctan(1/2 sqrt(4^x-4))+C