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

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

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Answer 1 · 2016-06-21T10:08:09

$\int 2 ln (x - 5) d x = (2 x - 10) \cdot ln (x - 5) - 2 x + C$ $color(blue)(int 2 ln(x-5) dx=(2x-10)*ln(x-5)-2x +C)$

Explanation:

The given

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

Let $u = ln (x - 5)$ $u=ln (x-5)$
Let $d v = d x$ $dv=dx$
Let $v = x$ $v=x$
Let $d u = (\frac{1}{x - 5}) \cdot d x$ $du=(1/(x-5))*dx$
Using integration by parts

$\int u \cdot d v = u v - \int v \cdot d u$ $int u*dv=uv-int v*du$

$\int ln (x - 5) d x = x \cdot ln (x - 5) - \int \frac{x}{x - 5} d x$ $int ln(x-5) dx=x*ln(x-5)-int x/(x-5)dx$

$\int ln (x - 5) d x = x \cdot ln (x - 5) - \int \frac{x - 5 + 5}{x - 5} d x$ $int ln(x-5) dx=x*ln(x-5)-int (x-5+5)/(x-5)dx$

$\int ln (x - 5) d x = x \cdot ln (x - 5) - \int (1 + \frac{5}{x - 5}) d x$ $int ln(x-5) dx=x*ln(x-5)-int (1+5/(x-5))dx$

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

So that

$\int 2 ln (x - 5) d x = 2 \cdot [x \cdot ln (x - 5) - x - 5 \cdot ln (x - 5)]$ $int 2 ln(x-5) dx=2*[x*ln(x-5)-x -5*ln(x-5)]$

$\int 2 ln (x - 5) d x = 2 x \cdot ln (x - 5) - 2 x - 10 \cdot ln (x - 5) + C$ $int 2 ln(x-5) dx=2x*ln(x-5)-2x -10*ln(x-5)+C$

$\int 2 ln (x - 5) d x = (2 x - 10) \cdot ln (x - 5) - 2 x + C$ $color(blue)(int 2 ln(x-5) dx=(2x-10)*ln(x-5)-2x +C)$

God bless....I hope the explanation is useful.

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

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