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:

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

$\blacksquare$

Examples

Primitive of $x^2 e^{-x}$

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


Also see


Sources

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