Primitive of x by Exponential of a x

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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

$\blacksquare$


Also presented as

This result is also seen presented in the form:

$\ds \int x e^{a x} \rd x = \frac {e^{a x} } {a^2} \paren {a x - 1} + C$


Also see


Sources

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