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

Integrals
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asked 2020-10-25
Evaluate the integral. $$\int_0^1\frac{(x-4)}{(x^2-5x+6dx)}$$

## Answers (1)

2020-10-26
$$\int_0^1\frac{x-4}{x^2-5x+6}dx$$
$$\int_0^1\frac{x-4}{x^2-3x-2x+6}dx$$
$$\int_0^1\frac{x-4}{[x^2-3x]+[-2x+6]}dx$$
$$\int_0^1\frac{x-4}{x[x-3x]-2[x-3]}dx$$
$$\int_0^1\frac{x-4}{(x-2)(x-3)}dx$$ We have,
$$\frac{x-4}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}$$
$$\frac{x-4}{(x-2)(x-3)}=\frac{A(x-3)}{(x-2)(x-3)}+\frac{B(x-2)}{(x-2)(x-3)}$$
$$\frac{x-4}{(x-2)(x-3)}=\frac{A(x-3)+B(x-2)}{(x-2)(x-3)}$$
On comparing the numerators, we get
$$x-4=A(x-3)+B(x-2)\Rightarrow$$ (1)
Substitute x=2 in Equ(1), To get
$$2-4=A(2-3)+B(2-2)$$
$$-2=-A\Rightarrow A=2$$
$$x-4=A(x-3)+B(x-2)\Rightarrow$$ (1)
Substitute x=3 in Eqn(1), To get
$$3-4=A(3-3)+B(3-2)$$
$$-1=B$$
$$\int_0^1\frac{x-4}{(x-2)(x-3)}dx=\int_0^1\frac{2}{x-2}-\frac{1}{x-3}dx$$
$$=[2\ln|x-2|-\ln|x-3|]_0^1$$
$$=[2\ln|1-2|-\ln|1-3|]-[2\ln|0-2|-\ln|0-3|]$$
$$=[0-\ln2]-[2\ln2-\ln3]=\ln3-3\ln2$$
$$=\ln3-\ln2^3-\ln8=\ln(\frac{3}{8})$$
Remember that: $$\ln x-\ln y=\ln(\frac{x}{y})$$
Also Remember that:$$a\cdot\ln x=\ln x^a$$
Result
$$\int_0^1\frac{x-4}{x^2-5x+6}dx=\ln3-3\ln2=\ln(\frac{3}{8})$$

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