Solve ∫(x+2)√(x-3)dx

  • Thread starter cunhasb
  • Start date
  • #1
12
0
Hoping anyone could give a hand on this....

this is what i've gotten so far...

∫(x+2)√(x-3)dx
u=(x-3)^1/2
u^2=(x-3)
x=u^2-3
dx=2u du

∫(x+2)√(x-3)dx=(u^2-3+2)*(u)*(2u) du =2∫(u^2-1)*(u)*(u)
= 2∫(u^2-1)*(u^2)
=2∫(u^4-u^2)
=(2*1/5*u^5) - ((2/5)*2*(1/3)*u^3)
=2/5*(x-3)^(5/2) - 4/15(x-3)^3/2 + K

Is it right?

Thank you so much guys...
 

Answers and Replies

  • #2
Tide
Science Advisor
Homework Helper
3,076
0
It would be much easier to write x + 2 as x - 3 + 5 and proceed from there.
 
  • #3
hotvette
Homework Helper
996
5
One sure way of checking yourself is to differentiate the result and see if you come with the original integrand.
 
  • #4
12
0
∫(x+2)√(x-3)dx
∫(x-3+5)√(x-3)dx

u=(x-3)^1/2
u^2=(x-3)
x=u^2-3
dx=1/du

∫(x+2)√(x-3)dx=(u^2+5)*(u^2) du
= 5∫(u^4) du +∫(u^2)
=5/5*u^5 + 1/3*u^3
=(x-3)^(5/2) - 1/3(x-3)^3/2 + K

Is it right now?

Thank you very much...


Thank you so much guys...
 
  • #5
12
0
I think the last one is wrong... here is the one I think might be right...
Can you please tell me if it is, and if not I would really appreciate if you could show me the way... thanks....


∫(x+2)√(x-3)dx
∫(x-3+5)√(x-3)dx

u=(x-3)^1/2
u^2=(x-3)
x=u^2-3
dx=2u

∫(x+2)√(x-3)dx=(u^2+5)*(u)*(2u) du
= 2∫(u^4+ 5u^2)
= 2∫(u^4 du +5∫u^2)
= 2*1/5*d^5 + 2*5*1/3*u^3
= 2/5*u^5 + 10/3*u^3
= 2/5(x-3)^(3/2) + 10/3(x-3)^(3/2) + K


Is it right now?

Thank you very much...


Thank you so much guys...
 
  • #6
12
0
I think I finally got this....


∫(x+2)√(x-3)dx
∫(x-3+5)√(x-3)dx

u=(x-3)^1/2
u^2=(x-3)
x=u^2-3
dx=2u

∫(x+2)√(x-3)dx=(u^2+5)*(u)*(2u) du
= 2∫(u^4+ 5u^2)
= 2∫(u^4 du +5∫u^2)
= 2*1/5*u^5 + 2*5*1/3*u^3
= 2/5*u^5 + 10/3*u^3
u*3(2/5u^2 +10/3) + k
(x-3)^(3/2) * (2/5(x-3) + 10/3) + K
(x-3)^(3/2) * (2/5x - 6/5 + 10/3) + k
(x-3)^(3/2) * (2/5x - 18/15 + 50/3) + k
(x-3)^(3/2) * (2/5x + 32/15) + k

Is it right now?

Thank you so much guys...
 
  • #7
VietDao29
Homework Helper
1,423
2
It looks correct. But you seem to make the problem more complicated. Here's a shorter one:
[tex]\int (x + 2) \sqrt{x - 3} dx = \int (x - 3 + 5) \sqrt{x - 3} dx = \int \left( (x - 3) ^ {\frac{3}{2}} + 5 (x - 3) ^ {\frac{1}{2}} \right) dx[/tex]
[tex]\int (x - 3) ^ {\frac{3}{2}} dx + \ \int 5 (x - 3) ^ {\frac{1}{2}} dx = \int (x - 3) ^ {\frac{3}{2}} d(x - 3) + \ 5 \int (x - 3) ^ {\frac{1}{2}} d(x - 3)[/tex]
[tex]= \frac{2}{5}(x - 3) ^ {\frac{5}{2}} + \frac{10}{3} (x - 3) ^ {\frac{3}{2}} + C[/tex]
-------------
cunhasb said:
∫(x+2)√(x-3)dx=(u^2+5)*(u)*(2u) du
= 2∫(u^4+ 5u^2)
= 2∫(u^4 du +5∫u^2)
= 2*1/5*u^5 + 2*5*1/3*u^3
= 2/5*u^5 + 10/3*u^3
Or you can just stop here and substitute u in:
[tex]u ^ 2 = x - 3 \Rightarrow u = \sqrt{x - 3}[/tex]
[tex]u ^ 5 = (x - 3) ^ {\frac{5}{2}}[/tex]
[tex]u ^ 3 = (x - 3) ^ {\frac{3}{2}}[/tex]
Viet Dao,
 

Related Threads on Solve ∫(x+2)√(x-3)dx

  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
20
Views
2K
Replies
6
Views
2K
Replies
2
Views
3K
Replies
9
Views
3K
Replies
1
Views
1K
Replies
4
Views
1K
Replies
5
Views
2K
Top