Javascript Disabled Detected

You currently have javascript disabled. Several functions may not work. Please re-enable javascript to access full functionality.

0

Integration

Started by AM4R, Dec 19 2006 09:20 PM

You cannot reply to this topic
Go to first unread post

2 replies to this topic

#1 AM4R

Site Swot

Members
145 posts

Gender:Male
Gender:Male

Posted 19 December 2006 - 09:20 PM

Can somone tell me where I am going wrong here becuase when I integrate this I dont get any 'ln' in the answer but the back of my textbook does???

x^2 + 3 / x

I get (x^2/2) +3

Thanks

::::::/\M/\R::::::

Back to top

#2 ad absurdum

Top of the Class

Members
390 posts

Location:Cambridge
Interests:Muzak.
Gender:Male

Posted 19 December 2006 - 09:39 PM

I'm afraid I can't see how you got your answer. This one isn't too tricky though, it should be just integrating each term.
$\int x^2 + \frac{3}{x} dx$ = $\int x^2 dx + \int \frac{3}{x} dx$

Integrating x^2

is higher stuff. To integrate $\frac{3}{x}$ you will need to use advanced higher knowledge (this should be where your natural log turns up).

HMFC - Founded 1874, beefing the Cabbage since 1875

Back to top

#3 euler

Newbie

Members
2 posts

Gender:Male

Posted 24 February 2007 - 05:38 PM

QUOTE(AM4R @ Dec 19 2006, 09:20 PM)

Can somone tell me where I am going wrong here becuase when I integrate this I dont get any 'ln' in the answer but the back of my textbook does???

x^2 + 3 / x

I get (x^2/2) +3

Thanks

$\begin{align*}\int \left(x^2 + \frac3x\right)\,dx=\frac13 x^3+3\ln{x}+C\end{align*}$

Back to top

Back to Mathematics

1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users