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∫((6x+3)/(x^2 + x)) dx All work is greatly appreciated
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3(log(x)+log(x+1) + c
oh no, the entire solution disapeared
It was wrong, sorry. When I replaced dx in the integral I accidentally used du.
oh, okay thanks! You really understand these concepts well
Here is a correct solution: \[ u=3(x^2+x) \\ du=(6x+3)dx \] and \[ \int{\frac{6x+3}{x^2+x}dx} = \int{\frac{du}{u/3}} = 3\int{\frac{du}{u}} = 3\ln{u}+C \] Substituting back u gives: \[ 3\ln{3(x^2+x)}+C = 3\ln(3x(x+1))+C = 3(\ln(3x)+\ln(x+1))+C \]
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But ln(3x) is just ln(3)+ln(x) and ln(3) can be included in the constant. So kryptons solution is equivalent.
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