How do you integrate $int xln x^3 dx$ using integration by parts?

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Answer 1 · 2016-12-12T20:36:49

$I=(3x^2)/4[2lnx-1]+C$

formula is: $" "intu(dv)/(dx)dx=uv-intv(du)/(dx)dx$

the choice of $" "u" "$ & $" "v$ is vital.

if logs are involved then it is usual to take $u=lnf(x)$ but try and simplify the expression using the laws of logs first if possible.

$intxlnx^3dx=int3xlnxdx$

$u=lnx=>(du)/(dx)=1/x$

$(dv)/(dx)=3x=>v=(3x^2)/2$

$I=(3x^2)/2lnx-int(3x^2)/2xx1/xdx$

$I=(3x^2)/2lnx-int(3x)/2dx$

$I=(3x^2)/2lnx-(3x^2)/4+C$

$I=(3x^2)/4[2lnx-1]+C$

How do you integrate int xln x^3 dx int xln x^3 dx using integration by parts?