How do you find the indefinite integral of $\int (\frac{3}{x}) d x$ $int (3/x)dx$ ?

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Answer 1 · 2016-11-11T16:52:02

$\int (\frac{3}{x}) d x = 3 ln x + C = ln (A x^{3})$ $int(3/x)dx=3lnx +C = ln(Ax^3)$

You should remember a standard special case:

$\frac{d}{d x} ln x = \frac{1}{x} \Leftrightarrow \int \frac{1}{x} d x = ln x + C$ $d/dxlnx = 1/x <=> int 1/xdx = lnx + C$

Hence, $\int (\frac{3}{x}) d x = 3 ln x + C$ $int(3/x)dx=3lnx +C$

NB If we write $C = ln A$ $C=lnA$ we can also write the solution as

$\int (\frac{3}{x}) d x = 3 ln x + ln A$ $int(3/x)dx=3lnx + lnA$
$:. int(3/x)dx=lnx^3 + lnA$
$:. int(3/x)dx=ln(Ax^3)$

How do you find the indefinite integral of int (3/x)dx∫(3x)dxint (3/x)dx?