Primitive of Exponential of a x over Power of x
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Theorem
- $\ds \int \frac {e^{a x} \rd x} {x^n} = \frac {-e^{a x} } {\paren {n - 1} x^{n - 1} } + \frac a {n - 1} \int \frac {e^{a x} \rd x} {x^{n - 1} } + C$
where $n \ne 1$.
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:
\(\ds u\) | \(=\) | \(\ds e^{a x}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds a e^{a x}\) | Derivative of $e^{a x}$ |
and let:
\(\ds \frac {\d v} {\d x}\) | \(=\) | \(\ds \frac 1 {x^n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^{-n}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds v\) | \(=\) | \(\ds \frac {x^{-n + 1} } {-n + 1}\) | Primitive of Power | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-1} {\paren {n - 1} x^{n - 1} }\) | simplifying |
Then:
\(\ds \int \frac {e^{a x} \rd x} {x^n}\) | \(=\) | \(\ds e^{a x} \paren {\frac {-1} {\paren {n - 1} x^{n - 1} } } - \int \paren {\frac {-1} {\paren {n - 1} x^{n - 1} } } \paren {a e^{a x} } \rd x + C\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-e^{a x} } {\paren {n - 1} x^{n - 1} } + \frac a {n - 1} \int \frac {e^{a x} \rd x} {x^{n - 1} } + C\) | Primitive of Constant Multiple of Function |
$\blacksquare$
Also see
- Primitive of $\dfrac {e^{a x} } x$ for the case where $n = 1$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $e^{a x}$: $14.514$