What is the integral of nxe^{nθ}e^{-xn} ?

The integral of nxe^{nθ}e^{-xn} is (e^{θn}(-ne^{-nx}x-e^{-nx}))/n+C

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integral of nxe^{nθ}e^{-xn}

\int n x e^{n θ} e^{- x n} d x

\frac{e ^{θ n} ( - n e ^{- n x} x - e ^{- n x} )}{n} + C

Solution steps

\int n x e^{n θ} e^{- x n} d x

Take the constant out:

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

= e^{θ n} n \cdot \int e^{- n x} x d x

Apply u-substitution

: \int \frac{e ^{u} u}{n ^{2}} d u

= e^{θ n} n \cdot \int \frac{e ^{u} u}{n ^{2}} d u

Take the constant out:

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

= e^{θ n} n \frac{1}{n ^{2}} \cdot \int e^{u} u d u

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

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

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

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

Substitute back

u = - n x

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

Simplify

e^{θ n} n \frac{1}{n ^{2}} (e^{- n x} (- n x) - e^{- n x}) : \frac{e ^{θ n} ( - n e ^{- n x} x - e ^{- n x} )}{n}

= \frac{e ^{θ n} ( - n e ^{- n x} x - e ^{- n x} )}{n}

Add a constant to the solution

= \frac{e ^{θ n} ( - n e ^{- n x} x - e ^{- n x} )}{n} + C

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What is the integral of nxe^{nθ}e^{-xn} ?
The integral of nxe^{nθ}e^{-xn} is (e^{θn}(-ne^{-nx}x-e^{-nx}))/n+C