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Calculating the integral of a function of the form $x^n \\ln(ax)$ using integration by parts.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Calculate the integral \\[ \\simplify{int(x^{n}ln({a}x),x)}\\]
", "advice": "If we have a function of $x$ which is the product of two functions of $x$, to integrate such a function it is often necessary to use Integration by Parts. The formula for Integration by Parts is:
\n\\[ \\int u(x) \\frac{dv}{dx} dx = u(x)v(x) - \\int v(x) \\frac{du}{dx} dx.\\]
\nUsing this method can be broken down into steps:
\nFor the integral
\n\\[ \\simplify{int(x^{n}ln({a}x),x)},\\]
\nwe must first identify $u(x)$ and $\\tfrac{dv}{dx}$. In this case, let \\[ u(x)=\\simplify{ln({a}x)},\\quad \\frac{dv}{dx}= \\simplify{x^{n}}. \\]
\nNext, we need to calculate $\\tfrac{du}{dx}$ and $v(x)$:
\n\\[ \\begin{split} u(x) = \\simplify{ln({a}x)} \\quad &\\implies \\frac{du}{dx} = \\frac{1}{x}; \\\\ \\frac{dv}{dx} = \\simplify{x^{n}} &\\implies v(x) = \\simplify{x^{n+1}/{n+1}}. \\end{split} \\]
\nPlugging these 4 terms into the integration by parts formula:
\n\\[ \\begin{split} \\simplify{int(x^{n}ln({a}x),x)} &\\,= \\simplify[fractionNumbers]{1/{n+1} ln({a}x)x^{n+1}} - \\simplify[fractionNumbers,alwaysTimes]{int((1/x) * (x^{n+1}/{n+1}),x)}, \\\\ \\\\ &\\,= \\simplify[fractionNumbers]{1/{n+1} ln({a}x) x^{n+1} - 1/{n+1} int(x^{n},x)}, \\\\ \\\\ &\\,=\\simplify[fractionNumbers]{1/{n+1}ln({a}x)x^{n+1} - 1/{(n+1)^2}x^{n+1}}+c. \\end{split} \\]
\nFinally, simplifying our solution:
\n\\[ \\simplify{int(x^{n}ln({a}x),x)} = \\simplify[fractionNumbers]{1/{(n+1)^2}x^{n+1}*[{n+1}ln({a}x)-1]}+c. \\]
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