How do you integrate int sqrtt lnttlnt by integration by parts method?

1 Answer
Nov 17, 2016

intsqrt(t)lntdt = 2/9t^(3/2)(3lnt-2) + C tlntdt=29t32(3lnt2)+C

Explanation:

If you are studying maths, then you should learn the formula for Integration By Parts (IBP), and practice how to use it:

intu(dv)/dxdx = uv - intv(du)/dxdx udvdxdx=uvvdudxdx, or less formally intudv=uv-intvdu udv=uvvdu

I was taught to remember the less formal rule in word; "The integral of udv equals uv minus the integral of vdu". If you struggle to remember the rule, then it may help to see that it comes a s a direct consequence of integrating the Product Rule for differentiation.

Essentially we would like to identify one function that simplifies when differentiated, and identify one that simplifies when integrated (or is at least is integrable).

So for the integrand sqrt(t)lnttlnt, hopefully you can see that lntlnt simplifies when differentiated and sqrt(t)t [effectively remains unchanged under differentiation or integration] [becomes harder to integrate]

Let { (u=lnt, => , (du)/dt=1/t), ((dv)/dt=t^(1/2), =>, v=2/3t^(3/2) ) :}

Then plugging into the IBP formula gives us:
int(u)((dv)/dx)dx = (u)(v) - int(v)((du)/dx)dx
:. int(lnt)(u^(1/2))dt = (lnt)(2/3t^(3/2)) - int(2/3t^(3/2))(1/t)dt
:. intsqrt(t)lntdt = 2/3t^(3/2)lnt - 2/3intu^(1/2)dt
:. intsqrt(t)lntdt = 2/3t^(3/2)lnt - 2/3t^(3/2)/(3/2) + C
:. intsqrt(t)lntdt = 2/3t^(3/2)lnt - 4/9t^(3/2) + C
:. intsqrt(t)lntdt = 2/9t^(3/2)(3lnt-2) + C