How do you integrate by substitution #int (t+2t^2)/sqrtt dt#?

Question

2625 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-26T23:06:10

#int(t+2t^2)/sqrt(t)dt=2sqrt(t^3)/3+4sqrt(t^5)/5+C#

We see a #sqrt(t)# our first thought is to substitude #u=sqrt(t)#.

So let's do that :

#u=sqrt(t)=>du=1/(2sqrt(t))dt=>dt=2sqrt(t)du=>dt=2udu#

So our integral becomes :

#int(t+2t^2)/sqrt(t)dt=int(u^2+2u^4)/u*2udu=2int(u^2+2u^4)du=#

#2(u^3/3+2u^5/5)+C=2u^3/3+4u^5/5+C#

Now let's substitude #u=sqrt(t)# back in :

#int(t+2t^2)/sqrt(t)dt=2sqrt(t^3)/3+4sqrt(t^5)/5+C#