Question

How do I solve this indefinite integral? ∫(3x^5-2x^4+5x^3+π)/(3x^4 )

Answer 1

You can split the fraction up and have 4 integrals

Answer 2

that seems unnecessary, just divide out the 3x^4 and it should be easy

Answer 3

first divide by 3x^4:

= 2/3 + 5x^-1 + 3pi x^-4
∫( 2/3 + 5x^-1 + 3pi x^-4) dx

= (2/3) x + 5 ln x + 3pi (x^(-3) / -3
= (2/3) x + 5 ln x - pi / x^3 + C

Answer 4

@joewhit you forgot about the 3x^5 term at the beginning

Answer 5

oh soi i did !

Answer 6

first divide by 3x^4:

= x+2/3 + 5x^-1 + 3pi x^-4
∫(x+ 2/3 + 5x^-1 + 3pi x^-4) dx
now integrate

Answer 7

then you have to add x^2 / 2 to my above result

Answer 8

Prgm
Local i,plural,clockwas,t,k,wait
ClrIO
"s"->plural
0->k
isClkOn()->clockwas
Define wait()=Func:EndFunc
ClockOn
For i,99,1,-1
Disp ""
Disp string(i)&" bottle"&plural&" of beer on the"
Disp "wall, "&string(i)&" bottle"&plural&" of beer."
getTime()[3]->t
While getTime()[3]=t and k=0:getKey()->k:EndWhile
if k != 0: exit
Disp "Take one down, pass it"
Disp "around."
getTime()[3]->t
While getTime()[3]=t and k=0:getKey()->k:EndWhile
If k!=0:Exit
If i-1=1:""->plural
If i>1 Then
Disp string(i-1)&" bottle"&plural&" of beer on the"
Disp "wall."
Else
Disp "No more bottles of beer on"
Disp "the wall"
EndIf
getTime()[3]->t
While getTime()[3]=t and k=0:getkey()->k:EndWhile
If k!=0:Exit
EndFor
If not clockwas:ClockOff
EndPrgm

Answer 9

adgdgdgd
LOL did you program the beer bottle car song?

Answer 10

wow - what language is that ? basic?

Answer 11

∫(x+ 2/3 + 5x^-1 + 3pi x^-4) dx
=x^2/2+(2/3) x + 5 ln x - pi / x^3 + C
(just to put it all in one place)

Answer 12

That looks like the language used for the TI calculator programs

Answer 13

good one - agdg should give you a beer

Answer 14

But when you divide , don't you get x - 2/3, instead of x +2/3??

Answer 15

yeah - you r right - 2/3

Answer 16

Looks like its basic

Answer 17

ok, here it is (hopefully) without errors
∫(x- 2/3 + 5x^-1 + 3pi x^-4) dx
=x^2/2-(2/3) x + 5 ln x - pi / x^3 + C

Answer 18

lol - finally!!

Answer 19

medals for all!

Answer 20

yeah!!!

Answer 21

Thank you so much

Answer 22

i think it is (x^2)/2 - (2/3)x + (5/3)lnx - (1/18x^3) (pi^2) +C

i think someone messed up

Answer 23

tmauldin, I think you're right!

Answer 24

but how do you get -(1/18^3)(pi^2) ??

Answer 25

ok ok sorry pi is a constant that gets factored out wt (1/3)
instead it is -pi/(9x^3)

Answer 26

i am 100 % positive the answer is (x^2)/2 - (2/3)x + (5/3)lnx - pi/(9x^3) +C
....if you have any doubts i can verify and asnwer any ?

Answer 27

tmauldin, you're a genious! XD

Answer 28

Your answer is right

Answer 29

no genious here, a genious wouldnt mess up the first try. glad i could help, hopefully ill remember pi is a constant..u helped me as well it was two ways

Answer 30

Wow, we made so many mistakes doing this problem. It wasn't even a hard one, we just weren't careful enough: We missed a term at the beginning, lost a negative, and failed to divide pi by 3 (somehow we got pi/3x^4=3pi/x^4...absurd). I guess we all need to be more careful with these silly details, and remember that just because we can do calculus doesn't mean we're very smart. Or at least very observant.