What is the antiderivative of #ln(x)#?

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Answer 1 · 2015-03-05T21:18:53

#I = intln(x)dx#

Let #ln(x) = t#

#=> x=e^t#

#=> dx = e^tdt#

Substituting in the Integral,

#I = intte^tdt#

On integrating by parts, keeping the first function as #t# and second function as #e^t#, we get

#I = tinte^tdt - int(dt/dtinte^tdt)dt#

Which is, simply,

#I = te^t - e^t + C#

#=> I = e^t ( t-1) + C#

Substituting the value of #t = ln(x)#,

#intln(x)dx = x[ln(x) - 1] + C#