# Primitive of x squared by Exponential of a x

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## Theorem

$\ds \int x^2 e^{a x} \rd x = \frac {e^{a x} } a \paren {x^2 - \frac {2 x} a + \frac 2 {a^2} } + C$

## 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:

 $\ds u$ $=$ $\ds x^2$ $\ds \leadsto \ \$ $\ds \frac {\d u} {\d x}$ $=$ $\ds 2 x$ Derivative of Power

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:

 $\ds \int x e^{a x} \rd x$ $=$ $\ds x^2 \paren {\frac {e^{a x} } a} - \int 2 x \frac {e^{a x} } a \rd x + C$ Integration by Parts $\ds$ $=$ $\ds x^2 \paren {\frac {e^{a x} } a} - \frac 2 a \int x e^{a x} \rd x + C$ Primitive of Constant Multiple of Function $\ds$ $=$ $\ds x^2 \paren {\frac {e^{a x} } a} - \frac 2 a \paren {\frac {e^{a x} } a \paren {x - \frac 1 a} } + C$ Primitive of $x e^{a x}$ $\ds$ $=$ $\ds \frac {e^{a x} } a \paren {x^2 - \frac {2 x} a + \frac 2 {a^2} } + C$ simplifying

$\blacksquare$