Answer:
Step-by-step explanation:
1.
To write the form of the partial fraction decomposition of the rational expression:
We have:
[tex]\mathbf{\dfrac{8x-4}{x(x^2+1)^2}= \dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}}[/tex]
2.
Using partial fraction decomposition to find the definite integral of:
[tex]\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}dx[/tex]
By using the long division method; we have:
[tex]x^2-8x-20 | \dfrac{2x}{2x^3-16x^2-39x+20 }[/tex]
[tex]- 2x^3 -16x^2-40x[/tex]
[tex]x+ 20[/tex]
So;
[tex]\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x+\dfrac{x+20}{x^2-8x-20}[/tex]
By using partial fraction decomposition:
[tex]\dfrac{x+20}{(x-10)(x+2)}= \dfrac{A}{x-10}+\dfrac{B}{x+2}[/tex]
[tex]= \dfrac{A(x+2)+B(x-10)}{(x-10)(x+2)}[/tex]
x + 20 = A(x + 2) + B(x - 10)
x + 20 = (A + B)x + (2A - 10B)
Now; we have to relate like terms on both sides; we have:
A + B = 1 ; 2A - 10 B = 20
By solvong the expressions above; we have:
[tex]A = \dfrac{5}{2}[/tex] [tex]B = \dfrac{3}{2}[/tex]
Now;
[tex]\dfrac{x+20}{(x-10)(x+2)} = \dfrac{5}{2(x-10)} + \dfrac{3}{2(x+2)}[/tex]
Thus;
[tex]\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)}[/tex]
Now; the integral is:
[tex]\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = \int \begin {bmatrix} 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)} \end {bmatrix} \ dx[/tex]
[tex]\mathbf{\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = x^2 + \dfrac{5}{2}In | x-10|\dfrac{3}{2} In |x+2|+C}[/tex]
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