Primitive of x by Square of Logarithm of x

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Theorem

$\ds \int x \ln^2 x \rd x = \dfrac {x^2 \ln^2 x} 2 - \dfrac {x^2 \ln x} 2 + \dfrac {x^2} 4 + C$


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:

\(\ds u\) \(=\) \(\ds x \ln x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds x \cdot \dfrac \d {\d x} \ln x + \ln x \dfrac \d {\d x} x\) Product Rule for Derivatives
\(\ds \) \(=\) \(\ds x \cdot \frac 1 x + \ln x \cdot 1\) Derivative of Natural Logarithm, Derivative of Identity Function
\(\ds \) \(=\) \(\ds 1 + \ln x\) simplifying


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds \ln x\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds x \ln x - x\) Primitive of $\ln x$


Then:

\(\ds \int x \ln^2 x \rd x\) \(=\) \(\ds \int \paren {x \ln x} \paren {\ln x} \rd x\)
\(\ds \) \(=\) \(\ds \paren {x \ln x} \paren {x \ln x - x} - \int \paren {x \ln x - x} \paren {1 + \ln x} \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds x^2 \ln^2 x - x^2 \ln x - \int x \ln x \rd x + \int x \rd x - \int x \ln^2 x \rd x + \int x \ln x \rd x + C\) Linear Combination of Primitives
\(\ds \leadsto \ \ \) \(\ds 2 \int x \ln^2 x \rd x\) \(=\) \(\ds x^2 \ln^2 x - x^2 \ln x + \int x \rd x + C\) simplifying
\(\ds \) \(=\) \(\ds x^2 \ln^2 x - x^2 \ln x + \dfrac {x^2} 2 + C\) Primitive of $x$
\(\ds \leadsto \ \ \) \(\ds \int x \ln^2 x \rd x\) \(=\) \(\ds \dfrac {x^2 \ln^2 x} 2 - \dfrac {x^2 \ln x} 2 + \dfrac {x^2} 4 + C\) simplifying

$\blacksquare$


Sources

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