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The formula for integrating by parts is

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\\[ \\int u\\frac{dv}{dx} dx = uv - \\int v \\frac{du}{dx} dx. \\]

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We choose $u = \\simplify[std]{ln{c}x}$ and $\\displaystyle \\frac{dv}{dx} = \\simplify[std]{{a}x}$.

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So $\\displaystyle \\frac{du}{dx} = \\simplify[std]{1/x}$ and $\\displaystyle v = \\simplify[std]{({a}x^2/2)}$.

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Hence,
\\[ \\begin{eqnarray} \\int \\simplify[std]{({a}*x)*ln({c}*x)} dx &=& uv - \\int v \\frac{du}{dx} dx \\\\ &=& \\simplify[std]{(({a}*x^2)/2)*ln({c}*x) - Int(({a}*x/2),x)} \\\\ &=& \\simplify[std]{(({a}x^2)/2)*ln({c}*x) -({a}x^2/4) + C} \\end{eqnarray} \\]

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$I=\\displaystyle \\int \\simplify[std]{({a}x)*ln({c}x)} dx $

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The formula for integration by parts is

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\\[ \\int u\\frac{dv}{dx} dx = uv - \\int v \\frac{du}{dx} dx. \\]

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What is the most suitable choice for $u$ and $\\frac{dv}{dx}$?

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$u =\\;$[[0]]

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$\\frac{dv}{dx} =\\;$[[1]]

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Hence find:

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$\\frac{du}{dx} =\\;$[[0]]

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$v =\\;$[[1]]

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Hence find:

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$uv =\\;$[[0]]

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$\\int v\\frac{du}{dx}\\mathrm{d}x = \\;$[[1]]$+C$

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Do not input numbers as decimals, only as integers without the decimal point, or fractions

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Use the results from above to find:

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$I=\\displaystyle \\int \\simplify[std]{({a}x)*ln({c}x)} dx =  uv - \\int v \\frac{du}{dx} dx = \\;$[[0]]$+C$

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Input all numbers as fractions or integers and not decimals.

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Find the following indefinite integral.

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This question is scaffolded - i.e. it takes you through answering the question step by step.

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Input all numbers as fractions or integers and not decimals.

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Don't forget $C$!

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Find $\\displaystyle \\int (ax)\\ln(cx)\\; dx $

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