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antiderivative(18)/(5-x) = kln[5-x]+c Find the exact value of k that makes the antidifferentiation formula true.
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i got 18 but is wrong
@ jorourk3 \[\int\limits_{}^{}\frac{ 18 }{ 5-x }=-18\ln(x-5)+C\]=kln[5-x]+c seems like k=-1 ? yes or no?
im sorry its -18 :D
k would be -18 then by what your saying...wouldnt it? that is also absolute value signs around ln[x-5] btw...does that change it from what you are saying?
its correct! thanks! i dont understand why it is negative though?
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its negative bec you let u=5-x then du= -dx
or du/dx=-1
ok thanks....can you help me with a similar one please? :)
ok yw ..:D just post them again so that others can see them
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