(2x^2-3x+8)dx/(2x^3-4x^2+8x-1)

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Ans:
I = \int \frac{2x^{2}-3x+8}{2x^{3}-4x^{2}+8x-1}dx
We will proceed this question by partial fraction rule.
f(x) = 2x^{3}-4x^{2}+8x-1
Roots of f(x):
x_{1} = .1333
x_{2} = .93335+1.69707i
x_{3} = .93335-1.69707i
\frac{2x^{2}-3x+8}{2x^{3}-4x^{2}+8x-1} = \frac{a}{x-x_{1}} + \frac{b}{x-x_{2}}+\frac{c}{x-x_{3}}
We have values of a, b & c from here,
I = \int (\frac{a}{x-x_{1}}+\frac{b}{x-x_{2}}+\frac{c}{x-x_{3}})dx
I = aln|x-x_{1}| + bln|x-x_{2}| + cln|x-x_{3}| + constant
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty
3 months ago

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