X(logX-1) + C
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logx^2=2 2logx=2 logx=1 10^1=x x=10
log(100x) can be written as log100 + logx. This =2+logx
Using u-substitution (where u = sinx), you'll find the antiderivative to be 0.5*sin2x + C.
(2/3)*x^(3/2)
If in the real number universe, first k is to be >0, y=kx = exlog(k) the antiderivative of eax is eax/a so the antiderivative of Y is exlog(k) / log(k) = kx /log(k)