What is the integral of (2^{2x}-1)/(sqrt(2^x)) ?

The integral of (2^{2x}-1)/(sqrt(2^x)) is 1/(ln(2))(2/3*2^{(3x)/2}+2^{1/2 (2-x)})+C

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integral of (2^{2x}-1)/(sqrt(2^x))

\int \frac{2 ^{2 x} - 1}{2 ^{x}} d x

\frac{1}{ln ( 2 )} (\frac{2}{3} \cdot 2^{\frac{3 x}{2}} + 2^{\frac{1}{2} (2 - x)}) + C

Solution steps

\int \frac{2 ^{2 x} - 1}{2 ^{x}} d x

Apply u-substitution

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

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

Expand

\frac{u ^{2} - 1}{u u} : u - \frac{1}{u u}

= \frac{1}{ln ( 2 )} \cdot \int u - \frac{1}{u 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}{ln ( 2 )} (\int u d u - \int \frac{1}{u u} d u)

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

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

= \frac{1}{ln ( 2 )} (\frac{2}{3} u^{\frac{3}{2}} - (- \frac{2}{u}))

Substitute back

u = 2^{x}

= \frac{1}{ln ( 2 )} (\frac{2}{3} (2^{x})^{\frac{3}{2}} - (- \frac{2}{2 ^{x}}))

Simplify

\frac{1}{ln ( 2 )} (\frac{2}{3} (2^{x})^{\frac{3}{2}} - (- \frac{2}{2 ^{x}})) : \frac{1}{ln ( 2 )} (\frac{2}{3} \cdot 2^{\frac{3 x}{2}} + 2^{\frac{1}{2} (2 - x)})

= \frac{1}{ln ( 2 )} (\frac{2}{3} \cdot 2^{\frac{3 x}{2}} + 2^{\frac{1}{2} (2 - x)})

Add a constant to the solution

= \frac{1}{ln ( 2 )} (\frac{2}{3} \cdot 2^{\frac{3 x}{2}} + 2^{\frac{1}{2} (2 - x)}) + C

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What is the integral of (2^{2x}-1)/(sqrt(2^x)) ?
The integral of (2^{2x}-1)/(sqrt(2^x)) is 1/(ln(2))(2/3*2^{(3x)/2}+2^{1/2 (2-x)})+C