int frac{3x+4}{x^3-2x-4} dx = log f(x) + C Ri | vedclass

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(A) Given the integral $I = \int \frac{3x+4}{x^3-2x-4} dx = \log f(x) + C$. First,factorize the denominator: $x^3-2x-4 = (x-2)(x^2+2x+2)$. Using parti

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$\frac{1}{\sqrt{17}}$

Vedclass · Answer

(A) Given the integral $I = \int \frac{3x+4}{x^3-2x-4} dx = \log f(x) + C$. First,factorize the denominator: $x^3-2x-4 = (x-2)(x^2+2x+2)$. Using partial fractions: $\frac{3x+4}{(x-2)(x^2+2x+2)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+2x+2}$. Equating coefficients: $3x+4 = A(x^2+2x+2) + (Bx+C)(x-2)$. For $x=2$: $3(2)+4 = A(4+4+2) \Rightarrow 10 = 10A \Rightarrow A=1$. Comparing $x^2$ coefficients: $0 = A+B \Rightarrow B = -1$. Comparing constants: $4 = 2A - 2C \Rightarrow 4 = 2 - 2C \Rightarrow 2C = -2 \Rightarrow C = -1$. Thus,$I = \int \frac{1}{x-2} dx - \int \frac{x+1}{x^2+2x+2} dx$. $I = \log|x-2| - \frac{1}{2} \int \frac{2x+2}{x^2+2x+2} dx = \log|x-2| - \frac{1}{2} \log|x^2+2x+2| + C$. $I = \log \left( \frac{|x-2|}{\sqrt{x^2+2x+2}} \right) + C$. So,$f(x) = \frac{|x-2|}{\sqrt{x^2+2x+2}}$. $f(3) = \frac{|3-2|}{\sqrt{3^2+2(3)+2}} = \frac{1}{\sqrt{9+6+2}} = \frac{1}{\sqrt{17}}$.

$\int \frac{3x+4}{x^3-2x-4} dx = \log f(x) + C \Rightarrow f(3) = ?$

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