integration help

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integration help

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|dw:1442933005067:dw|
so far this is what i have done|dw:1442933049561:dw|
thats not quite right, but you're on the right track with the substiution

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now im thinking of doing this let me know if im right|dw:1442933185892:dw|
@FireKat97 how would you solve it
|dw:1442933436297:dw|
wow Im lagging like crazy right now
looks ok 4x - 48 + 48
me 2 laggin so bad. @Empty i did a long division
\[\dfrac{4x}{x-12}=\dfrac{4x-48+48}{x-12} = 4 + \dfrac{48}{x-12}\]
@IrishBoy123 does my u and du ok and taking the constant 48 outside the integral a right move?
@ganeshie8 can u help out plz
rhe answer should be 4x+48ln(|x-12|)+c
oh one sec
omg I made the stupidest mistake ever lol its 1 am
:D
Ill write it up again hehe
alright step by step plz so i can understand how you solve it :D
sure :)
hang on, I keep getting 4x + 48ln(x -12) - 48 + C
but the -48 is a constant and so is C, so could they possibly have just replaced the (C -48) with a C?
because it is a constant.... omggg
|dw:1442935114878:dw|
Yup thats what I got too
this is how I did it incase you wanted to see
1 Attachment
wait no, the 48 isn't meant to multiply by the 4...
|dw:1442935301328:dw|
i think u break it into 2 different integral since we are adding
yeah that looks right
\[\int \dfrac{4x}{x-12}=\int \dfrac{4x-48+48}{x-12} \\ = \int 4 + \dfrac{48}{x-12} = \int 4 + \int \dfrac{48}{x-12} \\ = 4x + 48 ln(x-12) + C\] \[\checkmark\]
Thanks @IrishBoy123 and @firelord for the help.. really appreciated it
:)
yes but C - 48 = C', ie new constant and you get same answer
i got 2 more problems if you dont mind ill open a new thread and help me out :D

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