Primitive of Exponential of a x over Power of x

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Theorem

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

where $n \ne 1$.


Proof

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\displaystyle u\) \(=\) \(\displaystyle e^{a x}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d u} {\d x}\) \(=\) \(\displaystyle a e^{a x}\) Derivative of $e^{a x}$


and let:

\(\displaystyle \frac {\d v} {\d x}\) \(=\) \(\displaystyle \frac 1 {x^n}\)
\(\displaystyle \) \(=\) \(\displaystyle x^{-n}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {x^{-n + 1} } {-n + 1}\) Primitive of Power
\(\displaystyle \) \(=\) \(\displaystyle \frac {-1} {\paren {n - 1} x^{n - 1} }\) simplifying


Then:

\(\displaystyle \int \frac {e^{a x} \rd x} {x^n}\) \(=\) \(\displaystyle e^{a x} \paren {\frac {-1} {\paren {n - 1} x^{n - 1} } } - \int \paren {\frac {-1} {\paren {n - 1} x^{n - 1} } } \paren {a e^{a x} } \rd x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {-e^{a x} } {\paren {n - 1} x^{n - 1} } + \frac a {n - 1} \int \frac {e^{a x} \rd x} {x^{n - 1} } + C\) Primitive of Constant Multiple of Function

$\blacksquare$


Also see


Sources