one more i'm stuck on?

Question

probably yet another easy one, but once again i'm stuck.

take the integral from 1-->5 of
dx/(3x+1)

ok, i know you're supposed to use the substitution rule again and i'm saying u=3x+1. and then plugging it in, changing the upper/lower bounds, but it's not working...again. help?

Answer 1

∫ dx / (3x + 1)
Let, u = 3x +1 , du = 3 dx
∫[du/3u ]
= (1/3) ln(u)
= (1/3) ln(3x + 1)

1/3 [ln(16) - ln(4)]
= 1/3 [ln(4)]

Answer 2

The integral is 1/3*ln(3x+1) (eval at 1 and 5)

Let u = 3x+1 then du = 3dx and Integral is then integral(1/(3u)du) = 1/3ln(u) = 1/3ln(3x+1) = 1/3(ln(16) - ln(4))

Answer 3

i believe the integral is 1/3ln (3x +1)

use u = 3x +1
du = 3dx so dx = 1/3 du

then plug in 1 and 5

Answer 4

integral of 1/(3x+1) dx

let u = 3x + 1. then du = 3dx. so dx = (1/3) du

integral of (1/u) (1/3) du = (1/3) integral of (1/u) du = (1/3) ln (u) = (1/3) ln (3x + 1)

Evaluating...
(1/3) ln (16) - (1/3) ln (4) = 0.46

Answer 5

u = 3x + 1
du = 3dx
When x = 1, u = 4.
When x = 5, u = 16.
int(1 to 5) dx / (3x + 1)
= int(4 to 16) du / 3u
= (1/3) [ ln (u) ] (4 to 16)
= (1/3) (ln(16) - ln(4))
= (1/3) ln(16/4)
= ln(4) / 3.

Answer 6

int [dx / (3x + 1) ]
Let, u = 3x +1 , du = 3 dx
int [ du/3u ]
= (1/3) ln(u)
= (1/3) ln(3x + 1)
Now substitute the limits,
1/3 [ln(16) - ln(4)]
= 1/3 [ln(4)]