How do i integrate 1/(9x+1) in terms of x?

Answer 1

let u = 9x+1
let du=9dx

Now we have 1/9 integral of 9dx/(9x+1)
If we sub in u and du we have 1/9 integral of du/u
This is a natural log problem.
so we now have 1/9(lnu+c)
sub back in the value of u

the answer is:

1/9*ln I9x+1I +c

Note: If you don't remember your c and your absolute value notation, your answer will be wrong.

Answer 2

by substitution..

Let y = 9x + 1

it follows that dy = 9 dx

then integrate

1 / y * dx

or 1/9 * 1/y dy

= 1/9 * ln (y)

= 1/9 * ln (9x + 1)

Answer 3

If you differentiate ln (9x), you get (1/9x)(9)
If you differentiate ln (9x-1), you get (1/(9x+1))(9)
So, if you integrate, your answer shall be (1/9) ln (9x+1)

Answer 4

Solution:
z=9*x +1 ; dz = 9*dx
int (1/(9*x+1), dx) = int(1/z,dz/9) = (1/9)*ln(z) =1/9*ln(9*x+1)
Ans: 1/9*ln(9*x+1)

Cheers.

Answer 5

It is 1/9 LnI9x+1I