Primitive of x squared by Exponential of x/Proof 2

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Theorem

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


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:

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


Then:

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

$\blacksquare$


Sources

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