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


3 years ago

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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 & RegardsJitender SinghIIT DelhiaskIITians Faculty

one year ago

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