`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`
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