What is the integral of 1/((x*(e^2)^x)) ?

The integral of 1/((x*(e^2)^x)) is \Ei(-2x)+C

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integral of 1/((x*(e^2)^x))

integral

\frac{1}{( x \cdot ( e ^{2} ) ^{x} )}

Ei (- 2 x) + C

Solution steps

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

(e^{2})^{x} = e^{2 x}

= \int \frac{1}{e ^{2 x} x} d x

Apply u-substitution

: \int \frac{1}{e ^{u} u} d u

= \int \frac{1}{e ^{u} u} d u

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{e ^{u} u} = e^{- u} \frac{1}{u}

= \int e^{- u} \frac{1}{u} d u

Simplify

e^{- u} \frac{1}{u} : \frac{e ^{- u}}{u}

= \int \frac{e ^{- u}}{u} d u

Apply u-substitution

: \int \frac{e ^{v}}{v} d v

= \int \frac{e ^{v}}{v} d v

Use the nonelementary integral:

\int \frac{e ^{x}}{x} d x = Ei (x)

= Ei (v)

Substitute back

= Ei (- 2 x)

Add a constant to the solution

= Ei (- 2 x) + C

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

\frac{sin ^{4} ( x )}{( 1 + cos ( x ) ) ^{2}}

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

\frac{2}{x ^{2} - 9}

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

What is the integral of 1/((x*(e^2)^x)) ?
The integral of 1/((x*(e^2)^x)) is \Ei(-2x)+C