How do you integrate #int t^2(t^3+4)^(-1/2)# using substitution?

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Answer 1 · 2018-04-03T22:02:23

The integral is #2/3sqrt(t^3+4)+C#

First, rewrite the integral:

#color(white)=int t^2(t^3+4)^(-1/2)dt#

#=int t^2/(t^3+4)^(1/2)dt#

#=int t^2/sqrt(t^3+4)dt#

Now, let:

#u=t^3+4quadcolor(blue)=>quaddu=3t^2dtquadcolor(blue)=>quaddt=(du)/(3t^2)#

Substituting:

#=int t^2/sqrt(u)*(du)/(3t^2)#

#=int color(red)cancelcolor(black)(t^2)/sqrt(u)*(du)/(3color(reD)cancelcolor(black)(t^2))#

#=int 1/sqrt(u)*(du)/3#

#=1/3int1/sqrtudu#

#=1/3int1/u^(1/2)du#

#=1/3intu^(-1/2)du#

#=1/3*u^(-1/2+1)/(-1/2+1)#

#=1/3*u^(1/2)/(1/2)#

#=1/3*2*sqrtu#

#=2/3sqrtu#

Put #t^3+4# back in for #u# (and don't forget to add #C#):

#=2/3sqrt(t^3+4)+C#

That's it. Hope this helped!