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

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Answer 1 · 2015-03-05T21:18:53

$I = \int ln (x) d x$ $I = intln(x)dx$

Let $ln (x) = t$ $ln(x) = t$

$\Rightarrow x = e^{t}$ $=> x=e^t$

$\Rightarrow d x = e^{t} d t$ $=> dx = e^tdt$

Substituting in the Integral,

$I = \int t e^{t} d t$ $I = intte^tdt$

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

$I = t \int e^{t} d t - \int (\frac{d t}{d t} \int e^{t} d t) d t$ $I = tinte^tdt - int(dt/dtinte^tdt)dt$

Which is, simply,

$I = t e^{t} - e^{t} + C$ $I = te^t - e^t + C$

$\Rightarrow I = e^{t} (t - 1) + C$ $=> I = e^t ( t-1) + C$

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

$\int ln (x) d x = x [ln (x) - 1] + C$ $intln(x)dx = x[ln(x) - 1] + C$

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