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Mathematics
OpenStudy (anonymous):

Integration Question in comment, i don't know how to attach images in a question

OpenStudy (anonymous):

OpenStudy (anonymous):

oooops, i didn't mean dy/dx, i meant the integration sign

OpenStudy (anonymous):

aah - i thought that might be the case

OpenStudy (anonymous):

if there was the correct notification, would that be correct?

OpenStudy (anonymous):

split this up into partial fractions we get INT (1 - 7 ) dx ----- 3(x+3)

OpenStudy (anonymous):

can you continue from there?

OpenStudy (anonymous):

you rresult is not correct

OpenStudy (anonymous):

I don't understand what you have done, where has this 7 and this 3(x+3) came from?

OpenStudy (anonymous):

sorry guy - i copied the question down wrongly - i'll hav e another look at it

OpenStudy (anonymous):

ok, i was a bit worried when a 7 randomly appeared lol thanks for helping me out, im currently revising for an exam :D

OpenStudy (anonymous):

kind of a guess: is it this instead?

OpenStudy (anonymous):

or...

OpenStudy (anonymous):

first do long division to obtain INT 1 / 3(3x + 1) + INT 1/3 = 1/3 INT 1/ (3x+1) + x/3

OpenStudy (anonymous):

dividing 3x + 2 by 9x+ 3 gives 1/3 + 1 / (9x+ 3)

OpenStudy (anonymous):

am i right?

OpenStudy (anonymous):

must admit my integration is pretty rusty

OpenStudy (anonymous):

lol, mine too, and i have an exam on it in about 36 hours xD

OpenStudy (anonymous):

I thought that the question would involve natural logs when integrating (like when 1/x is integrated) doesn't it?

OpenStudy (anonymous):

yes - so integrating 1/3 INT 1/ (3x+1) = 1/9 ln| 3x+1| so finally answer is x/3 + 1/9 ln| 3x+1| + C

OpenStudy (anonymous):

yea thats correct - i checked it out on wolframalpha

OpenStudy (anonymous):

ok, thanks for your help :D

OpenStudy (anonymous):

yw

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