How do you find the integral of #sqrt (t)*ln (t) dt#?

1 Answer
Daan J.
Sep 4, 2015

use integration by parts to find #intsqrt(t)ln(t)dt=2/3t^(3/2)(ln(t)-2/3)+C#

Explanation:

We use integration by parts by noting that #sqrt(t)=d/dt(2/3t^(3/2))#.
Therefore
#intsqrt(t)ln(t)dt=2/3intln(t)d(t^(3/2))=2/3(ln(t)t^(3/2)-intt^(3/2)dln(t))#.
Since #d/dt(ln(t))=1/t# this gives
#intsqrt(t)ln(t)dt=2/3(ln(t)t^(3/2)-intt^(3/2)/tdt)=2/3(ln(t)t^(3/2)-intt^(1/2)dt)=2/3(ln(t)t^(3/2)-2/3t^(3/2))+C=2/3t^(3/2)(ln(t)-2/3)+C#
With C the integration constant.

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