Question

$integral \: of \: \frac{1}{x} \: is \: \: log(x)$

$why \: integral \: of \: \frac{1}{x - 2} \: is \: equal \: to \: log \: | x - 2 |$
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Answer 1

Given:

$\sf\displaystyle \int {\dfrac{1}{x} } \, dx=log(x)$

To find:

$\sf\displaystyle \int {\dfrac{1}{x-2} } \, dx$

Explanation:

$\sf Let \: x-2=t$

$\textsf{On differentiating both the sides, we get:}$

$\Rightarrow \sf d\dfrac{(x-2)}{dx}=\dfrac{dt}{dx}$

$\Rightarrow \sf d\dfrac{x}{dx}+ d\dfrac{2}{dx}=\dfrac{dt}{dx}$

$\Rightarrow \sf 1+0=\dfrac{dt}{dx}$

$\Rightarrow \sf 1=\dfrac{dt}{dx}$

$\Rightarrow \sf dx=1dt$

$\Rightarrow \sf dx=dt$

$\textsf{Substituting the value of x-2 in the question, we get:}$

$\Rightarrow\sf\displaystyle \int {\dfrac{1}{t } \, dt$

$\Rightarrow\sf log(u)+C$

$\textsf{On substituting the value of u, we get:}$

$\Rightarrow\sf log(x-2)+C$

$\Rightarrow\boxed{\sf log|x-2|+C}$

Hence, the  ∫1/x-2 = log|x-2|+C where C is some constant.