Solve ∫(x+2)√(x-3)dx

1. Oct 1, 2005

cunhasb

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...

2. Oct 1, 2005

Tide

It would be much easier to write x + 2 as x - 3 + 5 and proceed from there.

3. Oct 2, 2005

hotvette

One sure way of checking yourself is to differentiate the result and see if you come with the original integrand.

4. Oct 2, 2005

cunhasb

∫(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. Oct 2, 2005

cunhasb

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. Oct 2, 2005

cunhasb

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. Oct 2, 2005

VietDao29

It looks correct. But you seem to make the problem more complicated. Here's a shorter one:
$$\int (x + 2) \sqrt{x - 3} dx = \int (x - 3 + 5) \sqrt{x - 3} dx = \int \left( (x - 3) ^ {\frac{3}{2}} + 5 (x - 3) ^ {\frac{1}{2}} \right) dx$$
$$\int (x - 3) ^ {\frac{3}{2}} dx + \ \int 5 (x - 3) ^ {\frac{1}{2}} dx = \int (x - 3) ^ {\frac{3}{2}} d(x - 3) + \ 5 \int (x - 3) ^ {\frac{1}{2}} d(x - 3)$$
$$= \frac{2}{5}(x - 3) ^ {\frac{5}{2}} + \frac{10}{3} (x - 3) ^ {\frac{3}{2}} + C$$
-------------
Or you can just stop here and substitute u in:
$$u ^ 2 = x - 3 \Rightarrow u = \sqrt{x - 3}$$
$$u ^ 5 = (x - 3) ^ {\frac{5}{2}}$$
$$u ^ 3 = (x - 3) ^ {\frac{3}{2}}$$
Viet Dao,