How do I find the integral $\int \frac{x}{x - 6} d x$ $intx/(x-6)dx$ ?

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Answer 1 · 2014-09-30T02:00:46

By Substitution,

$\int \frac{x}{x - 6} d x = x + 6 ln | x - 6 | + C$ $int x/{x-6}dx=x+6ln|x-6|+C$

Let us look at some details.

$\int \frac{x}{x - 6} d x$ $int x/{x-6}dx$

by the sunstitution $u = x - 6$ $u=x-6$ ,
$\Rightarrow x = u + 6$ $Rightarrow x=u+6$
$\Rightarrow d x = d u$ $Rightarrow dx=du$

$= \int \frac{u + 6}{u} d u$ $=int {u+6}/u du$

by splitting the integrand,

$= \int (1 + \frac{6}{u}) d u$ $=int (1+6/u) du$

$= u + 6 ln | u | + C_{1}$ $=u+6ln|u|+C_1$

by putting $u = x - 6$ $u=x-6$ back in,

$= x - 6 + 6 ln | x - 6 | + C_{1}$ $=x-6+6ln|x-6|+C_1$

by letting $C = C_{1} - 6$ $C=C_1-6$ ,

$= x + 6 ln | x - 6 | + C$ $=x+6ln|x-6|+C$

I hope that this was helpful.

How do I find the integral intx/(x-6)dx∫xx−6dxintx/(x-6)dx ?