What is #f(x) = int 1/(x+3) dx# if #f(2) = 0 #?

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Answer 1 · 2016-12-07T18:02:32

The function is #f(x) = ln|x + 3| - ln5#

We can rewrite this as #f(x) = int(x + 3)^-1dx#.

Let #u = x + 3#. Then #du = 1(dx) -> dx= du#.

#f(x) = int(u^-1)dx#

#f(x) = ln|u| + C#

#f(x) = ln|x+ 3| + C#

We can now solve for #C#. When #x = 2#, #y= 0#.

#0 = ln|2 + 3| + C#

#C = -ln5#

#:.# Therefore, the function is #f(x) = ln|x + 3| - ln5#.

Hopefully this helps!