How do you integrate $2 ln (x - 5)$ $2ln(x-5)$ ?

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Answer 1 · 2015-06-05T21:34:21

$2 \int 1 \cdot ln (x - 5) d x$ $2int 1*ln(x-5)dx$

You can integrate this using integration by parts. The general formula for integration by parts is:

$u v - \int v d u$ $uv - intvdu$

Now for w-substitution, let
$x - 5 = w$ $x-5 = w$
$x = w + 5$ $x = w+5$
$d w = d x$ $dw = dx$

$2 \int ln (x - 5) d x = 2 \int ln w d w$ $2intln(x-5)dx = 2intlnwdw$

For integration by parts, let
$u = ln w$ $u = lnw$
$d v = 1 d w$ $dv = 1dw$
$v = w$ $v = w$
$d u = \frac{1}{w} d w$ $du = 1/wdw$

$= 2 [w ln | w | - \int \frac{w}{w} d w]$ $= 2[wln|w| - intw/wdw]$

$= 2 [w ln | w | - w + C]$ $= 2[wln|w| - w + C]$

$= 2 [(x - 5) ln | x - 5 | - (x - 5) + C]$ $= 2[(x-5)ln|x-5| - (x-5) + C]$

$= 2 [x ln | x - 5 | - 5 ln | x - 5 | - x + 5 + C]$ $= 2[xln|x-5| - 5ln|x-5| - x + 5 + C]$

But $5$ $5$ is a constant, so it gets embedded into $C$ $C$ . Multiplying $C$ $C$ by $2$ $2$ still gives $C$ $C$ .

$= 2 [x ln | x - 5 | - 5 ln | x - 5 | - x] + C$ $= 2[xln|x-5| - 5ln|x-5| - x] + C$

How do you integrate 2ln(x−5)2ln(x-5)?