Pure Mathematics - Integration

x=ln(t-1)^0.5
differentiate both side with respect to t
LHS=dx/dt
RHS=[1/(t-1)^0.5]*(t-1)^-0.5*1/2=1/[2(t-1)]
therefore dx=1/[2(t-1)]*dt


then replace x in term of t in the question
dx=1/[2(t-1)]*dt

then the question become 1/2︴(t-1)t^1/3 dt
=1/2︴t^4/3 dt - 1/2︴t^1/3dt
done



︴=integration

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