How do I find the integral $\int x^{5} \cdot ln (x) d x$ $intx^5*ln(x)dx$ ?

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How do I find the integral $\int x^{5} \cdot ln (x) d x$ $intx^5*ln(x)dx$ ?

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Answer 1 · 2014-09-24T04:19:36

By Integration by Parts,

$\int x^{5} ln x d x = \frac{x^{6}}{36} (6 ln x - 1) + C$ $int x^5lnx dx=x^6/36(6lnx-1)+C$

Let us look at some details.

Let $u = ln x$ $u=lnx$ and $d v = x^{5} d x$ $dv=x^5dx$ .
$\Rightarrow d u = \frac{d x}{x}$ $Rightarrow du={dx}/x$ and $v = \frac{x^{6}}{6}$ $v=x^6/6$

By Integration by Parts

$\int u d v = u v - \int v d u$ $int udv=uv-int vdu$ ,

we have

$\int (ln x) \cdot x^{5} d x = (ln x) \cdot \frac{x^{6}}{6} - \int \frac{x^{6}}{6} \cdot \frac{d x}{x}$ $int (lnx)cdot x^5dx=(lnx)cdot x^6/6-int x^6/6cdot dx/x$

by simplifying a bit,

$= \frac{x^{6}}{6} ln x - \int \frac{x^{5}}{6} d x$ $=x^6/6lnx-int x^5/6dx$

by Power Rule,

$= \frac{x^{6}}{6} ln x - \frac{x^{6}}{36} + C$ $=x^6/6lnx-x^6/36+C$

by factoring out $\frac{x^{6}}{36}$ $x^6/36$ ,

$= \frac{x^{6}}{36} (6 ln x - 1) + C$ $=x^6/36(6lnx-1)+C$

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How do I find the integral $\int x^{5} \cdot ln (x) d x$ $intx^5*ln(x)dx$ ?

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