What is the integral of 2^{e^x}(1+2^{e^x})^5e^x ?

The integral of 2^{e^x}(1+2^{e^x})^5e^x is ((1+2^{e^x})^6)/(6ln(2))+C

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integral of 2^{e^x}(1+2^{e^x})^5e^x

\int 2^{e^{x}} (1 + 2^{e^{x}})^{5} e^{x} d x

\frac{( 1 + 2 ^{e^{x}} ) ^{6}}{6 ln ( 2 )} + C

Solution steps

\int 2^{e^{x}} (1 + 2^{e^{x}})^{5} e^{x} d x

Apply u-substitution

: \int 2^{u} (1 + 2^{u})^{5} d u

= \int 2^{u} (1 + 2^{u})^{5} d u

Apply u-substitution

: \int \frac{v ^{5}}{ln ( 2 )} d v

= \int \frac{v ^{5}}{ln ( 2 )} 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 \int v^{5} d v

Apply the Power Rule

: \frac{v ^{6}}{6}

= \frac{1}{ln ( 2 )} \cdot \frac{v ^{6}}{6}

Substitute back

= \frac{1}{ln ( 2 )} \cdot \frac{( 1 + 2 ^{e^{x}} ) ^{6}}{6}

Simplify

\frac{1}{ln ( 2 )} \cdot \frac{( 1 + 2 ^{e^{x}} ) ^{6}}{6} : \frac{( 1 + 2 ^{e^{x}} ) ^{6}}{6 ln ( 2 )}

= \frac{( 1 + 2 ^{e^{x}} ) ^{6}}{6 ln ( 2 )}

Add a constant to the solution

= \frac{( 1 + 2 ^{e^{x}} ) ^{6}}{6 ln ( 2 )} + C

\frac{d}{d x} (\frac{sec ( x )}{x})

\frac{d}{d x} ((ln (3))^{x})

\int \frac{x}{2} d x

f (x) = 3 x^{2} + 9 x^{3} - 5 x, at x = - 4

\frac{1}{2 ( s + 2 )}

What is the integral of 2^{e^x}(1+2^{e^x})^5e^x ?
The integral of 2^{e^x}(1+2^{e^x})^5e^x is ((1+2^{e^x})^6)/(6ln(2))+C