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

Question

Samantha Adkison · Accepted Answer

#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#

Charles Arocha · Answer

undefined To integrate the expression ∫(t + 2t^2)/√t dt by substitution:

1. Let u = √t.
2. Find du/dt and dt in terms of du.
3. Rewrite the integral in terms of u.
4. Integrate with respect to u.
5. Substitute back in terms of t.

Let's proceed with the calculations:

1. Let u = √t.
   Then, u^2 = t.

2. Find du/dt:
   du/dt = (1/2) * (t)^(-1/2).
   Solving for dt:
   dt = 2u dt.

3. Rewrite the integral in terms of u:
   ∫(t + 2t^2)/√t dt = ∫(u^2 + 2u^4) * 2u du.

4. Integrate with respect to u:
   ∫(2u^2 + 4u^5) du = (2/3)u^3 + (4/6)u^6 + C,
   where C is the constant of integration.

5. Substitute back in terms of t:
   = (2/3)√t^3 + (2/3)t^3/2 + C.