Primitive of x by Exponential of a x

Theorem

$\ds \int x e^{a x} \rd x = \frac {e^{a x} } a \paren {x - \frac 1 a} + C$

Proof

With a view to expressing the primitive in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

 $\ds u$ $=$ $\ds x$ $\ds \leadsto \ \$ $\ds \frac {\d u} {\d x}$ $=$ $\ds 1$ Derivative of Identity Function

and let:

 $\ds \frac {\d v} {\d x}$ $=$ $\ds e^{a x}$ $\ds \leadsto \ \$ $\ds v$ $=$ $\ds \frac {e^{a x} } a$ Primitive of $e^{a x}$

Then:

 $\ds \int x e^{a x} \rd x$ $=$ $\ds x \paren {\frac {e^{a x} } a} - \int \frac {e^{a x} } a \rd x + C$ Integration by Parts $\ds$ $=$ $\ds x \paren {\frac {e^{a x} } a} - \frac 1 a \int e^{a x} \rd x + C$ Primitive of Constant Multiple of Function $\ds$ $=$ $\ds x \paren {\frac {e^{a x} } a} - \frac 1 a \paren {\frac {e^{a x} } a} + C$ Primitive of $e^{a x}$ $\ds$ $=$ $\ds \frac {e^{a x} } a \paren {x - \frac 1 a} + C$ simplifying

$\blacksquare$