# What is f(x) = int 1/(x-3)-x/(x-2) dx if f(-1)=6 ?

Feb 7, 2016

$- x + \ln \left(x - 3\right) - 2 \ln \left(x - 2\right) + 2 + c$

#### Explanation:

$\int \frac{1}{x - 3} - \frac{x}{x - 2} \mathrm{dx}$
$\ln \left(x - 3\right) - \int \frac{x}{x - 2} \mathrm{dx}$
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$\int \frac{x}{x - 2} \mathrm{dx} = \int x {\left(x - 2\right)}^{-} 1 \mathrm{dx}$
Via integration by parts
$x \ln \left(x - 2\right) - \int \ln \left(x - 2\right) \mathrm{dx}$

Let $y = \ln \left(x - 2\right)$ so ${e}^{y} = x - 2$ and ${e}^{y} \mathrm{dy} = \mathrm{dx}$
$x \ln \left(x - 2\right) - \int \ln \left(x - 2\right) \mathrm{dx} = x \ln \left(x - 2\right) - \int y {e}^{y} \mathrm{dy}$
Integrate by parts again.
$x \ln \left(x - 2\right) - \int \ln \left(x - 2\right) \mathrm{dx} = x \ln \left(x - 2\right) - y {e}^{y} + {e}^{y}$
$x \ln \left(x - 2\right) - \left(x - 2\right) \ln \left(x - 2\right) + x - 2$
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$\int \frac{1}{x - 3} - \frac{x}{x - 2} \mathrm{dx}$
$\ln \left(x - 3\right) - \int \frac{x}{x - 2} \mathrm{dx}$
$\ln \left(x - 3\right) - x \ln \left(x - 2\right) + \left(x - 2\right) \ln \left(x - 2\right) - x + 2$
Simplify:
$f \left(x\right) = \ln \left(x - 3\right) - 2 \ln \left(x - 2\right) - x + 2 + C$
Substitute $f \left(- 1\right) = 6$