What is the integral of (e^e)^x*e^2x ?

The integral of (e^e)^x*e^2x is e^{ex+1}x-e^{ex}+C

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integral of (e^e)^x*e^2x

integral

(e^{e})^{x} \cdot e^{2} x

e^{e x + 1} x - e^{e x} + C

Solution steps

\int (e^{e})^{x} e^{2} x d x

Take the constant out:

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

= e^{2} \cdot \int (e^{e})^{x} x d x

(e^{e})^{x} = e^{e x}

= e^{2} \cdot \int e^{e x} x d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

: u ln (u) - \int 1 d u

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

\int 1 d u = u

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

Substitute back

u = e^{e x}

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

Simplify

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

= e^{e x + 1} x - e^{e x}

Add a constant to the solution

= e^{e x + 1} x - e^{e x} + C

\frac{( 1 - e ^{- x} )}{x}

\frac{sin ( x )}{cos ^{3} ( x )}

\frac{2 x}{( 1 - x ^{2} )}

ln (3 x - 2)

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

What is the integral of (e^e)^x*e^2x ?
The integral of (e^e)^x*e^2x is e^{ex+1}x-e^{ex}+C