substitution doesn't seem to work. integration by parts? partial fractions?
2007-02-01 03:15:05 · 5 answers · asked by Fred 1 in Science & Mathematics ➔ Mathematics
int x/(3x+2) dx = int 1/3 - 2/3(3x+2) dx
use u = 3x + 2 for the second term. This gives du = 3 dx and 2/9 * ln(3x+2) for the antiderivative.
final answer x/3 - 2 ln|3x+2| / 9 + C
2007-02-01 03:20:37 · answer #1 · answered by Ken M 3 · 0⤊ 0⤋
The easiest way for me is
x/(3x+2)= 1/3 x/(x+2/3) = 1/3 (x+2/3-2/3) /( x+2/3)= 1/3((1-2/3/(x+2/3))
Integrating now Int = 1/3 x -2/9 Ln Ix+2/3I +C
2007-02-01 11:44:50 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋
Divide top line by bottom line to get:-
1/3 - (2/3)/(3x + 2) = 1/3 - (2/9)3/(3x +2)
I = â« (1/3 dx - 2/9 â« 3/(3x + 2) dx = (1/3)x - 2/9(log(3x + 2) + C
2007-02-01 11:47:44 · answer #3 · answered by Como 7 · 0⤊ 0⤋
_1/3__
x / 3x+2
x+2/3
----------
-2/3
therefore
int 1/3-2/3(3x+2) dx= x/3-2/9ln[3x+2]+c
though i dont get the right ans. w/ substitution if sumbody can tell me
2007-02-01 11:29:45 · answer #4 · answered by Maths Rocks 4 · 0⤊ 0⤋
if you make t=3x+2 then x=(t-2)/3 and dt=3dx
then you'll have to integrate (t-2)/9t dt and it is easier
the answer is (3x+2)/9 - 2/9*ln(3x+2)
2007-02-01 11:59:33 · answer #5 · answered by Eddy 1 · 1⤊ 0⤋