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find the integral

Mathematics
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i attempted to solve it but i can't get the answer.. which is 24
\[\frac{1}{3}\ln(x^3+3x)\]

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Other answers:

just guessed
yeah i got up to there but i can't get the answer :/
1st take x2=t nd find is dx/dy
its easy but i dont have pen/pencil let me try give few min.
\[\huge{\ln(x^2+3x)}\huge{|_{1}^{3}}={\ln(3^2+3(3))-\ln(1^2+3(1))}\]
sorry i forgot\[\frac{1}{3}\] on boh sides

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