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integral ln(3x+5)
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for most things with\[\int\ln(ax+b)dx\]you can use\[dv=dx\]\[u=\ln(ax+b)\]
@yabucoayankee you there?
okay thats why i couldnt get it. theen what?
so then du= (1/3x+5)*3
yes, du=3/(3x+5)dx and what is v ?
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v= x
right :) so\[\int udv=uv-\int vdu\]is what after these substitutions?
still with me?
ln(3x+5)(x)-\int(x)*(1/3x+5)*3\]
yes
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so now you need to integrate\[\int{3x\over3x+5}dx\]any idea how to do that?
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