What is $f (x) = \int \frac{1}{x - 4}$ $f(x) = int 1/(x-4)$ if $f (2) = 1$ $f(2)=1$ ?

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Answer 1 · 2018-06-21T05:45:01

$f (x) = ln | x - 4 | + ln (\frac{e}{2})$ $f(x)=ln|x-4|+ln(e/2)$

OR

$f (x) = ln | x - 4 | + 1 - ln 2$ $f(x)=ln|x-4|+1-ln2$

Here,

$f (x) = \int \frac{1}{x - 4} d x$ $f(x)=int1/(x-4)dx$

$f(x)=ln|x-4|+c....to(1)$

Given that ,

$f(2)=1$

$=>ln|2-4|+c=1$

$=>ln|-2|+c=1$

$=>c=1-ln2$

$=>c=lne-ln2...to[becauselne=1]$

$=>c=ln(e/2)$

Subst. $c=ln(e/2)$ , into $(1)$

$f(x)=ln|x-4|+ln(e/2)$

What is f(x) = int 1/(x-4) f(x)=∫1x−4f(x) = int 1/(x-4) if f(2)=1 f(2)=1f(2)=1 ?