int frac{dx}{1 + x + x^2 + x^3} = | vedclass

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Question

(A) We have the integral $I = \int \frac{dx}{1 + x + x^2 + x^3}$.Factor the denominator: $1 + x + x^2 + x^3 = (1 + x) + x^2(1 + x) = (1 + x)(1 + x^2)$

Vedclass · Accepted Answer

$\log \sqrt{1 + x} - \frac{1}{2} \log \sqrt{1 + x^2} + \frac{1}{2} 	an^{-1} x + c$

Vedclass · Answer

(A) We have the integral $I = \int \frac{dx}{1 + x + x^2 + x^3}$.Factor the denominator: $1 + x + x^2 + x^3 = (1 + x) + x^2(1 + x) = (1 + x)(1 + x^2)$.Using partial fractions: $\frac{1}{(1 + x)(1 + x^2)} = \frac{A}{1 + x} + \frac{Bx + C}{1 + x^2}$.$1 = A(1 + x^2) + (Bx + C)(1 + x)$.For $x = -1$,$1 = A(1 + 1) \implies A = \frac{1}{2}$.Comparing coefficients: $A + B = 0 \implies B = -\frac{1}{2}$,and $A + C = 1 \implies C = \frac{1}{2}$.So,$I = \int (\frac{1/2}{1 + x} + \frac{-1/2 x + 1/2}{1 + x^2}) dx = \frac{1}{2} \int \frac{dx}{1 + x} - \frac{1}{4} \int \frac{2x}{1 + x^2} dx + \frac{1}{2} \int \frac{dx}{1 + x^2}$.$I = \frac{1}{2} \log|1 + x| - \frac{1}{4} \log(1 + x^2) + \frac{1}{2} 	an^{-1} x + c$.$I = \log \sqrt{1 + x} - \frac{1}{2} \log \sqrt{1 + x^2} + \frac{1}{2} 	an^{-1} x + c$.

$\int \frac{dx}{1 + x + x^2 + x^3} = $

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