1. ## integration again...

i know this is integration by parts but not sure how to actually do it..

$integrate......... 2x sqrt(x) dx$

2. Originally Posted by b00yeah05
i know this is integration by parts but not sure how to actually do it..

$integrate......... 2x sqrt(x) dx$
Why do you want to use integration by parts?

$\int (2x\sqrt{x}) dx = \int (2 \sqrt{x^5}) dx = \int (2 \cdot x^{\frac52}) dx$

I'll leave the rest for you!

3. where did you get the 5 from??????

4. Originally Posted by b00yeah05
where did you get the 5 from??????
Sorry - it's too early in the morning here. Instead of 5 write 3. You should have then:

$\int (2 \cdot x^{\frac32}) dx$

5. sorry..would you mind explaining how you got that?? im assuming you did...

2 . x . sqrt(x) = 2 . sqrt(x^2)

then 2 . (x^2)^1/2

correct?

6. Originally Posted by b00yeah05
sorry..would you mind explaining how you got that?? im assuming you did...

2 . x . sqrt(x) = 2 . sqrt(x^2)

then 2 . (x^2)^1/2

correct?
no. $x \sqrt{x}$ is not $\sqrt{x^2}$ (how did you get that??)

$x \sqrt{x} = x \cdot x^{1/2} = x^{1 + 1/2} = x^{3/2}$

7. Originally Posted by b00yeah05
sorry..would you mind explaining how you got that?? im assuming you did...

2 . x . sqrt(x) = 2 . sqrt(x^2)

then 2 . (x^2)^1/2

correct?
Not quite:

$x = \sqrt{x^2}$

Therefore

$x \sqrt{x} = \sqrt{x^2} \cdot \sqrt{x} = \sqrt{x^2 \cdot x} = \sqrt{x^3} = x^{\frac32}$

8. i see so...the x at the front becomes x^1 = x . x^1/2 therefore......x^1+1/2 = x^3/2

9. Originally Posted by b00yeah05
i see so...the x at the front becomes x^1 = x . x^1/2 therefore......x^(1+1/2) = x^3/2
Good job - and much easier than the calculations I did.

But: Use parantheses!

10. Originally Posted by Krizalid
I think this topic it's pretty useless having this one which is also solved.
indeed!

11. Originally Posted by Krizalid
I think this topic it's pretty useless having this one which is also solved.
And, in a broader context, there's this one, where the elephant in the room actually gets pointed out.