int frac{dx}{x^3+3x^2+2x} = | vedclass

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(D) Let $I = \int \frac{dx}{x^3+3x^2+2x}$.Factorizing the denominator: $x^3+3x^2+2x = x(x^2+3x+2) = x(x+1)(x+2)$.Using partial fractions: $\frac{1}{x(

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$\frac{1}{2} \log \left(\frac{|x^2+2x|}{(x+1)^2}\right) + c$

Vedclass · Answer

(D) Let $I = \int \frac{dx}{x^3+3x^2+2x}$.Factorizing the denominator: $x^3+3x^2+2x = x(x^2+3x+2) = x(x+1)(x+2)$.Using partial fractions: $\frac{1}{x(x+1)(x+2)} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x+2}$.$1 = A(x+1)(x+2) + Bx(x+2) + Cx(x+1)$.For $x=0$: $1 = A(1)(2) \Rightarrow A = \frac{1}{2}$.For $x=-1$: $1 = B(-1)(1) \Rightarrow B = -1$.For $x=-2$: $1 = C(-2)(-1) \Rightarrow C = \frac{1}{2}$.Thus,$I = \int \left(\frac{1}{2x} - \frac{1}{x+1} + \frac{1}{2(x+2)}\right) dx$.$I = \frac{1}{2} \log |x| - \log |x+1| + \frac{1}{2} \log |x+2| + c$.$I = \frac{1}{2} [\log |x| - 2 \log |x+1| + \log |x+2|] + c$.$I = \frac{1}{2} \log \left|\frac{x(x+2)}{(x+1)^2}\right| + c = \frac{1}{2} \log \left|\frac{x^2+2x}{(x+1)^2}\right| + c$.

$\int \frac{dx}{x^3+3x^2+2x} = $

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