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

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

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

integral

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

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

Solution steps

\int e^{2} x (e^{e})^{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 x (e^{e})^{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{( 2 t ^{2} - 1 )}{t}

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

\frac{x ^{3}}{( x + 1 ) ^{3}}

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

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

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