Primitive of Power of x by Logarithm of x

From ProofWiki
Jump to navigation Jump to search

Theorem

$\displaystyle \int x^m \ln x \rd x = \frac {x^{m + 1} } {m + 1} \left({\ln x - \frac 1 {m + 1} }\right) + C$

where $m \ne -1$.


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:

\(\displaystyle u\) \(=\) \(\displaystyle \ln x\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\d u} {\d x}\) \(=\) \(\displaystyle \frac 1 x\) Derivative of $\ln x$


and let:

\(\displaystyle \frac {\d v} {\d x}\) \(=\) \(\displaystyle x^m\)
\(\displaystyle \implies \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {x^{m + 1} } {m + 1}\) Primitive of Power


Then:

\(\displaystyle \int x^m \ln x \rd x\) \(=\) \(\displaystyle \frac {x^{m + 1} } {m + 1} \ln x - \int \frac {x^{m + 1} } {m + 1} \left({\frac 1 x}\right) \rd x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^{m + 1} } {m + 1} \ln x - \frac 1 {m + 1} \int x^m \rd x + C\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^{m + 1} } {m + 1} \ln x - \frac 1 {m + 1} \left({\frac {x^{m + 1} } {m + 1} }\right) + C\) Primitive of Power
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^{m + 1} } {m + 1} \left({\ln x - \frac 1 {m + 1} }\right) + C\) simplifying

$\blacksquare$


Example

$\displaystyle \int_1^n x^m \ln x \rd x = \frac {n^{m + 1} } {m + 1} \left({\ln n - \frac 1 {m + 1} }\right) + \frac 1 {\left({m + 1}\right)^2}$

where $m \ne -1$.


Also see


Sources