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
$\displaystyle u' = 1$. Not sure how you found your du.
Your dv and v is fine.
$\displaystyle \dfrac{2}{3}x(x+1)^{3/2} - \int \dfrac{2}{3}(x+1)^{3/2} \cdot 1 \ dx$
If you integrate using the power rule you end up with $\displaystyle \dfrac{2}{3}x(x+1)^{3/2} - \dfrac{4}{15}(x+1)^{5/2} + C = \dfrac{2}{15}(x+1)^{3/2} \left(5x - 2(x+1)\right)$
$\displaystyle = \dfrac{2}{15}(x+1)^{3/2} (3x-2) + C$
I would make the substitution $\displaystyle \displaystyle u = x + 1 \implies x = u - 1$ and $\displaystyle \displaystyle du = dx$.
Then
$\displaystyle \displaystyle \begin{align*} \int{x(x + 1)^{\frac{1}{2}}\,dx} &= \int{(u - 1)u^{\frac{1}{2}}\,du} \\ &= \int{u^{\frac{3}{2}} - u^{\frac{1}{2}}\,du} \\ &= \frac{u^{\frac{5}{2}}}{\frac{5}{2}} - \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + C \\ &= \frac{2(x + 1)^{\frac{5}{2}}}{5} - \frac{2(x + 1)^{\frac{3}{2}}}{3} + C \end{align*}$
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???
I edited my post above which has the easiest method, using a u substitution, now that the LaTeX is fixed you should be able to see it.
If you have to use integration by parts:
$\displaystyle \displaystyle u = x \implies du = dx, dv = (x + 1)^{\frac{1}{2}}\,dx \implies v = \frac{2}{3}(x + 1)^{\frac{3}{2}} $, so
$\displaystyle \displaystyle \begin{align*} \int{x(x + 1)^{\frac{1}{2}}\,dx} &= \frac{2}{3}x(x + 1)^{\frac{3}{2}} - \int{\frac{2}{3}(x + 1)^{\frac{3}{2}}\,dx} \\ &= \frac{2}{3}x(x + 1)^{\frac{3}{2}} - \frac{2}{3}\int{(x + 1)^{\frac{3}{2}}\,dx} \\ &= \frac{2}{3}x(x + 1)^{\frac{3}{2}} - \frac{2}{3}\cdot \frac{2}{5}(x + 1)^{\frac{5}{2}} + C \\ &= \frac{2}{3}x(x + 1)^{\frac{3}{2}} - \frac{4}{15}(x + 1)^{\frac{5}{2}} + C \end{align*}$
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- $\displaystyle \int (2/3)(x+ 1)^{3/2}dx$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????
$\displaystyle = (2/3)\int u^{3/2}du= (2/3)(2/5)u^{5/2}= (4/15)(x+1)^{5/2}+ C$