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

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

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Answer 1 · 2016-04-19T22:13:52

$\int x ln x d x = \frac{x^{2} ln x}{2} - \frac{x^{2}}{4} + C$ $intxlnxdx=(x^2lnx)/2-x^2/4+C$

Explanation:

Using integration by parts:

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

In the case of

$\int x ln x d x$ $intxlnxdx$

We let

$u = ln x \Rightarrow \frac{d u}{d x} = \frac{1}{x} \Rightarrow d u = \frac{1}{x} d x$ $u=lnx" "=>" "(du)/dx=1/x" "=>" "du=1/xdx$

$d v = x d x \Rightarrow \int d v = \int x d x \Rightarrow v = \frac{x^{2}}{2}$ $dv=xdx" "=>" "intdv=intxdx" "=>" "v=x^2/2$

Thus, plugging these in, we see that

$\int x ln x d x = (\frac{x^{2}}{2}) ln x - \int \frac{x^{2}}{2} (\frac{1}{x}) d x$ $intxlnxdx=(x^2/2)lnx-intx^2/2(1/x)dx$

$= \frac{x^{2} ln x}{2} - \int \frac{x}{2} d x$ $=(x^2lnx)/2-intx/2dx$

$= \frac{x^{2} ln x}{2} - \frac{x^{2}}{4} + C$ $=(x^2lnx)/2-x^2/4+C$

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

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