Integration by substitution problem
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u=the square root of x+2
Then
x=u squared minus 2
then
(integrate) x/(sqrt(x+2)) dx=(integrate)2u(squared)-4 du=2u(cubed)/3 - 4u
then substituting u as the square root of x+2 gives you
2((x+2)to the 3/2 power) over 3 minus (4 times the square root of (x+2))
Sorry for being unable to write the steps mathematically clear for I cannot type in math notations in my keyboard.
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When you integrate the above, you're right with u=x+2.
(integrate) x/(sqrt(x+2)) dx
(integrate) x/sqrt(u) dx, where u=x+2 and du=dx.
So you'd get:
(integrate) x/sqrt(u) du
I think your question is what to do with the numerator. Keep in mind that you set u=x+2. Therefore, x must equal... u-2. Subtract 2 from both sides.
(integrate) (u-2)/sqrt(u) du
Split the numerator to get:
(integrate) u/sqrt(u) - 2/sqrt(u)
(integrate) u^(1/2) - 2u^(-1/2)
Then you'd get 2/3(u)^(3/2) - 4u^(1/2) + C.
Substituting back, you'd get 2/3(x+2)^(3/2) - 4(x+2)^(1/2) + C.
Hopefully I've answered your question! Feel free to ping me if you have any questions from my explanation.
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