What is the integral of re^{kr} ?

The integral of re^{kr} is 1/(k^2)(e^{kr}kr-e^{kr})+C

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integral of re^{kr}

\int r e^{k r} d r

\frac{1}{k ^{2}} (e^{k r} k r - e^{k r}) + C

Solution steps

\int r e^{k r} d r

Apply u-substitution

= \int \frac{e ^{u} u}{k ^{2}} d u

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

Substitute back

u = k r

= \frac{1}{k ^{2}} (e^{k r} k r - e^{k r})

Add a constant to the solution

= \frac{1}{k ^{2}} (e^{k r} k r - e^{k r}) + C

What is the integral of re^{kr} ?
The integral of re^{kr} is 1/(k^2)(e^{kr}kr-e^{kr})+C