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Calculating the integral of a function of the form $\\ln(ax)$ using integration by parts.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Calculate the integral \\[ \\simplify{int(ln({a}x) ,x)}\\]
", "advice": "To calculate the integral
\n\\[ \\simplify{int(ln({a}x),x)},\\] despite there only being a single function of $x$, we can make use of integration by parts. This is because there is no 'standard' integral for $\\simplify{ln({a}x)}$.
\nHowever, if we think of \\[ \\simplify{int(ln({a}x),x)} = \\int 1 \\times \\ln(\\var{a}x)\\, \\text{d}x ,\\]
\nwe can set \\[ u=\\simplify{ln({a}x)}, \\quad \\frac{dv}{dx}=1,\\]
\nso we only have to differentiate the $\\simplify{ln({a}x)}$ term.
\nNow, calculating the $\\tfrac{du}{dx}$ and $v(x)$ terms:
\n\\[ \\begin{split} u(x) = \\simplify{ln({a}x)} \\quad &\\implies \\frac{du}{dx} = \\simplify{1/x}; \\\\ \\frac{dv}{dx} = 1 &\\implies v(x) = x. \\end{split} \\]
\nPlugging these 4 terms into the integration by parts formula we can succesfully find a solution:
\n\\[ \\begin{split} \\simplify{int(ln({a}x),x)} &\\,= \\simplify{x ln({a}x)} - \\simplify[alwaysTimes]{int((1/x)x,x)}, \\\\ &\\,=\\simplify{x ln({a}x)-int(1,x)}, \\\\ &\\,= \\simplify{x ln({a}x) - x}+c, \\\\ &\\,=\\simplify{x (ln({a}x)-1)}+c.\\end{split} \\]
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