What is the antiderivative of $ln x$ $lnx$ ?

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Answer 1 · 2018-03-16T21:07:52

$\int ln x d x = x (ln x - 1) + c$ $intlnxdx=x(lnx-1)+"c"$

To find an antiderivative of $ln x$ $lnx$ , we must find $\int ln x d x$ $intlnxdx$ . To do so, we use integration by parts.

$\int u d v = u v - \int v d u$ $intudv=uv-intvdu$

Let $u = ln x \Rightarrow d u = \frac{1}{x} d x$ $u=lnx rArr du=1/xdx$

And $d v = d x \Rightarrow v = x$ $dv=dx rArr v=x$

So

$\int ln x d x = x ln x - \int d x = x ln x - x = x (ln x - 1)$ $intlnxdx=xlnx-intdx=xlnx-x=x(lnx-1)$

What is the antiderivative of lnxlnx?