What is the integral of (16^x)/(sqrt(4^x+2)) ?

The integral of (16^x)/(sqrt(4^x+2)) is 1/(ln(2))(1/3 (4^x+2)^{3/2}-2sqrt(4^x+2))+C

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integral of (16^x)/(sqrt(4^x+2))

\int \frac{1 6 ^{x}}{4 ^{x} + 2} d x

\frac{1}{ln ( 2 )} (\frac{1}{3} (4^{x} + 2)^{\frac{3}{2}} - 2 4^{x} + 2) + C

Solution steps

\int \frac{1 6 ^{x}}{4 ^{x} + 2} d x

Apply u-substitution

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

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

\int 2 d u = 2 u

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

Substitute back

u = 4^{x} + 2

= \frac{1}{ln ( 2 )} (\frac{( 4 ^{x} + 2 ) ^{3}}{3} - 2 4^{x} + 2)

Simplify

\frac{1}{ln ( 2 )} (\frac{( 4 ^{x} + 2 ) ^{3}}{3} - 2 4^{x} + 2) : \frac{1}{ln ( 2 )} (\frac{1}{3} (4^{x} + 2)^{\frac{3}{2}} - 2 4^{x} + 2)

= \frac{1}{ln ( 2 )} (\frac{1}{3} (4^{x} + 2)^{\frac{3}{2}} - 2 4^{x} + 2)

Add a constant to the solution

= \frac{1}{ln ( 2 )} (\frac{1}{3} (4^{x} + 2)^{\frac{3}{2}} - 2 4^{x} + 2) + C

What is the integral of (16^x)/(sqrt(4^x+2)) ?
The integral of (16^x)/(sqrt(4^x+2)) is 1/(ln(2))(1/3 (4^x+2)^{3/2}-2sqrt(4^x+2))+C