Primitive of Exponential of a x by Logarithm of x
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Theorem
- $\ds \int e^{a x} \ln x \rd x = \frac {e^{a x} \ln x} a - \frac 1 a \int \frac {e^{a x} } x \rd x + C$
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:
\(\ds u\) | \(=\) | \(\ds \ln x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds \frac 1 x\) | Derivative of Natural Logarithm |
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:
\(\ds \int e^{a x} \ln x \rd x\) | \(=\) | \(\ds \ln x \paren {\frac {e^{a x} } a} - \int \paren {\frac {e^{a x} } a} \frac 1 x \rd x + C\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {e^{a x} \ln x} a - \frac 1 a \int \frac {e^{a x} } x \rd x + C\) | Primitive of Constant Multiple of Function |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $e^{a x}$: $14.522$