Primitive of Exponential of a x over Power of x

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 $n = 1$

• Primitive of $x^n e^{a x}$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $e^{a x}$: $14.514$