What is the integral of 4^x(4^x+1)^3 ?

The integral of 4^x(4^x+1)^3 is ((4^x+1)^4)/(8ln(2))+C

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integral of 4^x(4^x+1)^3

\int 4^{x} (4^{x} + 1)^{3} d x

\frac{( 4 ^{x} + 1 ) ^{4}}{8 ln ( 2 )} + C

Solution steps

\int 4^{x} (4^{x} + 1)^{3} d x

Apply u-substitution

: \int \frac{u ^{3}}{2 ln ( 2 )} d u

= \int \frac{u ^{3}}{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}{2 ln ( 2 )} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

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

Substitute back

u = 4^{x} + 1

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

Simplify

\frac{1}{2 ln ( 2 )} \cdot \frac{( 4 ^{x} + 1 ) ^{4}}{4} : \frac{( 4 ^{x} + 1 ) ^{4}}{8 ln ( 2 )}

= \frac{( 4 ^{x} + 1 ) ^{4}}{8 ln ( 2 )}

Add a constant to the solution

= \frac{( 4 ^{x} + 1 ) ^{4}}{8 ln ( 2 )} + C

(- \frac{1}{2}, 1), (- 2, \frac{1}{2})

\frac{d}{d x} (sin (x^{3} + 2))

\frac{\partial}{\partial x} (4 x - 2)

x \to 0 + lim (tan^{x} (3 x))

\frac{d y}{d t} = 3 t + \frac{y}{2 t}

What is the integral of 4^x(4^x+1)^3 ?
The integral of 4^x(4^x+1)^3 is ((4^x+1)^4)/(8ln(2))+C