Calculus of a Single Variable, Chapter 8, 8.2, Section 8.2, Problem 33

dy/dt=t^2/sqrt(3+5t)
y=intt^2/sqrt(3+5t)dt
Apply integral substitution :u=sqrt(3+5t)
du=1/2(3+5t)^(1/2-1)(5)dt
du=5/(2sqrt(3+5t))dt
=>dt/sqrt(3+5t)=2/5du
u=sqrt(3+5t)
squaring above,
u^2=3+5t
=>5t=u^2-3
t=(u^2-3)/5
t^2=1/25(u^2-3)^2
t^2=1/25(u^4-6u^2+9)
y=intt^2/(sqrt(3+5t))dt
=int1/25(u^4-6u^2+9)(2/5)du
=int2/125(u^4-6u^2+9)du
Take the constant out,
=2/125int(u^4-6u^2+9)du
Apply the sum and power rule,
=2/125(intu^4du-int6u^2du+int9du)
=2/125(u^5/5-6u^3/3+9u)
=2/125(u^5/5-2u^3+9u)
Substitute back u=sqrt(3+5t)
=2/125(1/5(3+5t)^(5/2)-2(3+5t)^(3/2)+9(3+5t)^(1/2))
simplify the above and add a constant C to the solution,
=2/125(3+5t)^(1/2)(1/5(3+5t)^2-2(3+5t)+9)+C
=2/125sqrt(3+5t)(1/5(9+25t^2+30t)-2(3+5t)+9)+C
=2/125sqrt(3+5t)((9+25t^2+30t-10(3+5t)+45)/5)+C
=2/125sqrt(3+5t)((9+25t^2+30t-30-50t+45)/5)+C
=2/125sqrt(3+5t)((25t^2-20t+24)/5)+C
y=2/625(25t^2-20t+24)sqrt(3+5t)+C