How do I find the integral $intx^5*ln(x)dx$ ?

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Answer 1 · 2014-09-24T04:19:36

By Integration by Parts,

$int x^5lnx dx=x^6/36(6lnx-1)+C$

Let us look at some details.

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

By Integration by Parts

$int udv=uv-int vdu$ ,

we have

$int (lnx)cdot x^5dx=(lnx)cdot x^6/6-int x^6/6cdot dx/x$

by simplifying a bit,

$=x^6/6lnx-int x^5/6dx$

by Power Rule,

$=x^6/6lnx-x^6/36+C$

by factoring out $x^6/36$ ,

$=x^6/36(6lnx-1)+C$

How do I find the integral intx^5*ln(x)dxintx^5*ln(x)dx ?