Question

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

Integrals
ANSWERED
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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