How do you integrate #t^(1/2)*ln(t)#?

1 Answer
maganbhai P.
Mar 5, 2018

#I=2/9*t^(3/2)[3*lnt-2]+C#

Explanation:

Integration by Parts:
#color(red)(intu*vdt=uintvdt-int{(du)/(dt)intvdt}dt)#
Let,#u=lnt,and,v=t^(1/2)rArr(du)/(dt)=1/t,intvdt=t^((1/2)+1)/((1/2)+1)=(t(3/2))/(3/2)#
#I=intt^(1/2)*lntdt=lnt*intt^(1/2)dt-intd/(dt)(lnt)dt*intt^(1/2)dtrArrI=lnt*t^(3/2)/(3/2)-int1/t*t^(3/2)/(3/2)*dt=2/3*t^(3/2)lnt-2/3intt^(3/2-1)dt=2/3t^(3/2)*lnt-2/3intt^(1/2)dt=2/3t^(3/2)*lnt-2/3*t^(3/2)/(3/2)+C#
#I=2/3*t^(3/2)*lnt-4/9*t^(3/2)+C#
#I=2/9*t^(3/2)[3*lnt-2]+C#

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