Primitive of Power of x by Logarithm of x

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Theorem

$\ds \int x^m \ln x \rd x = \frac {x^{m + 1} } {m + 1} \paren {\ln x - \frac 1 {m + 1} } + C$

where $m \ne -1$.


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:

\(\ds u\) \(=\) \(\ds \ln x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \frac 1 x\) Derivative of $\ln x$


and let:

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


Then:

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