# Math Help - Did I do this correct, integration of parts

1. ## Did I do this correct, integration of parts

S x(x+1)^.5 dx

u = 1x v= 2/3(x+1)^3/2
du = 1 dv (x+1)^.5

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

2. ## Re: Did I do this correct, integration of parts

$u' = 1$. Not sure how you found your du.

Your dv and v is fine.

$\dfrac{2}{3}x(x+1)^{3/2} - \int \dfrac{2}{3}(x+1)^{3/2} \cdot 1 \ dx$

If you integrate using the power rule you end up with $\dfrac{2}{3}x(x+1)^{3/2} - \dfrac{4}{15}(x+1)^{5/2} + C = \dfrac{2}{15}(x+1)^{3/2} \left(5x - 2(x+1)\right)$

$= \dfrac{2}{15}(x+1)^{3/2} (3x-2) + C$

3. ## Re: Did I do this correct, integration of parts

OK, I got the problem with your help, but the (5-2(x+1)) shouldn't the 5 have a 5x, as the 2/3x has an x with it?? Or did I miss something?

4. ## Re: Did I do this correct, integration of parts

i think you are right.but anyway besides that everything else should be correct

5. ## Re: Did I do this correct, integration of parts

Ok just making sure, thanks again for your help. always appreciated with the help always.
Joanne

6. ## Re: Did I do this correct, integration of parts

OK, I got the problem with your help, but the (5-2(x+1)) shouldn't the 5 have a 5x, as the 2/3x has an x with it?? Or did I miss something?
Yes, there should be. Good spot!

7. ## Re: Did I do this correct, integration of parts

S x(x+1)^.5 dx

u = 1x v= 2/3(x+1)^3/2
du = 1 dv (x+1)^.5

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 would make the substitution $\displaystyle u = x + 1 \implies x = u - 1$ and $\displaystyle du = dx$.

Then

\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*}

8. ## Re: Did I do this correct, integration of parts

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???

9. ## Re: Did I do this correct, integration of parts

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:

$\displaystyle u = x \implies du = dx, dv = (x + 1)^{\frac{1}{2}}\,dx \implies v = \frac{2}{3}(x + 1)^{\frac{3}{2}}$, so

\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*}

10. ## Re: Did I do this correct, integration of parts

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????

11. ## Re: Did I do this correct, integration of parts

if u = x and du = 1.
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.

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????
The "1" doesn't become anything- $\int (2/3)(x+ 1)^{3/2}dx$
$= (2/3)\int u^{3/2}du= (2/3)(2/5)u^{5/2}= (4/15)(x+1)^{5/2}+ C$

12. ## Re: Did I do this correct, integration of parts

$\displaystyle du$ does NOT equal $\displaystyle 1, \frac{du}{dx} = 1$ and $\displaystyle du = dx$.