# Evaluate the integral. int_0^1frac{(x-4)}{(x^2-5x+6dx)}

Evaluate the integral. ${\int }_{0}^{1}\frac{\left(x-4\right)}{\left({x}^{2}-5x+6dx\right)}$
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${\int }_{0}^{1}\frac{x-4}{{x}^{2}-5x+6}dx$
${\int }_{0}^{1}\frac{x-4}{{x}^{2}-3x-2x+6}dx$
${\int }_{0}^{1}\frac{x-4}{\left[{x}^{2}-3x\right]+\left[-2x+6\right]}dx$
${\int }_{0}^{1}\frac{x-4}{x\left[x-3x\right]-2\left[x-3\right]}dx$
${\int }_{0}^{1}\frac{x-4}{\left(x-2\right)\left(x-3\right)}dx$ We have,
$\frac{x-4}{\left(x-2\right)\left(x-3\right)}=\frac{A}{x-2}+\frac{B}{x-3}$
$\frac{x-4}{\left(x-2\right)\left(x-3\right)}=\frac{A\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\frac{B\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}$
$\frac{x-4}{\left(x-2\right)\left(x-3\right)}=\frac{A\left(x-3\right)+B\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}$
On comparing the numerators, we get
$x-4=A\left(x-3\right)+B\left(x-2\right)⇒$ (1)
Substitute x=2 in Equ(1), To get
$2-4=A\left(2-3\right)+B\left(2-2\right)$
$-2=-A⇒A=2$
$x-4=A\left(x-3\right)+B\left(x-2\right)⇒$ (1)
Substitute x=3 in Eqn(1), To get
$3-4=A\left(3-3\right)+B\left(3-2\right)$
$-1=B$
${\int }_{0}^{1}\frac{x-4}{\left(x-2\right)\left(x-3\right)}dx={\int }_{0}^{1}\frac{2}{x-2}-\frac{1}{x-3}dx$
$=\left[2\mathrm{ln}|x-2|-\mathrm{ln}|x-3|{\right]}_{0}^{1}$
$=\left[2\mathrm{ln}|1-2|-\mathrm{ln}|1-3|\right]-\left[2\mathrm{ln}|0-2|-\mathrm{ln}|0-3|\right]$
$=\left[0-\mathrm{ln}2\right]-\left[2\mathrm{ln}2-\mathrm{ln}3\right]=\mathrm{ln}3-3\mathrm{ln}2$
$=\mathrm{ln}3-\mathrm{ln}{2}^{3}-\mathrm{ln}8=\mathrm{ln}\left(\frac{3}{8}\right)$
Remember that: $\mathrm{ln}x-\mathrm{ln}y=\mathrm{ln}\left(\frac{x}{y}\right)$
Also Remember that:$a\cdot \mathrm{ln}x=\mathrm{ln}{x}^{a}$
Result
${\int }_{0}^{1}\frac{x-4}{{x}^{2}-5x+6}dx=\mathrm{ln}3-3\mathrm{ln}2=\mathrm{ln}\left(\frac{3}{8}\right)$