S x(x+1)^.5 dx
u = 1x v= 2/3(x+1)^3/2
du = 1 dv (x+1)^.5
answer
2/3x(x+1)^3/2 -2(x+1)^3 + c
Let me know thanks.
CORRECTED MY du, I copied it down wrong, it should be 1
S x(x+1)^.5 dx
u = 1x v= 2/3(x+1)^3/2
du = 1 dv (x+1)^.5
answer
2/3x(x+1)^3/2 -2(x+1)^3 + c
Let me know thanks.
CORRECTED MY du, I copied it down wrong, it should be 1
I have another question. Tried this question again.
2/3(x)(x+1)^3/2 -2/3 int sign (x+1)^3/2 * 1 dx
So I get
2/3(x)(x+1)^3/2 - 2/5x(x+1)^5/2 +c
The question is the 1 above which is du, so the integral is x so I just pop it infront of the 2/5???
if u = x and du = 1.
well when you put the first line after the negative integral sign, it's 2/3(x+1)^3/2 * 1 dx. as du is 1?
When you take the integral of the 1 wouldn't it be x then the other becomes 2/5(x+1)^5/2????
You keep saying this even though the first response to your post told you it was wrong. If u= x, then du= dx, not 1.
The "1" doesn't become anything-well when you put the first line after the negative integral sign, it's 2/3(x+1)^3/2 * 1 dx. as du is 1?
When you take the integral of the 1 wouldn't it be x then the other becomes 2/5(x+1)^5/2????