If int frac{dx}{x^4+5x^2+4} = A tan^{-1} x + | vedclass

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Question

(A) Let $I = \int \frac{dx}{x^4+5x^2+4}$.Factor the denominator: $x^4+5x^2+4 = (x^2+1)(x^2+4)$.Using partial fractions: $\frac{1}{(x^2+1)(x^2+4)} = \f

Vedclass · Accepted Answer

$A = \frac{1}{3}, B = -\frac{1}{6}$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{x^4+5x^2+4}$.Factor the denominator: $x^4+5x^2+4 = (x^2+1)(x^2+4)$.Using partial fractions: $\frac{1}{(x^2+1)(x^2+4)} = \frac{A'}{x^2+1} + \frac{B'}{x^2+4}$.$1 = A'(x^2+4) + B'(x^2+1)$.Setting $x^2 = -1$,we get $1 = A'(3) \implies A' = \frac{1}{3}$.Setting $x^2 = -4$,we get $1 = B'(-3) \implies B' = -\frac{1}{3}$.So,$I = \int \left( \frac{1/3}{x^2+1} - \frac{1/3}{x^2+4} ight) dx$.$I = \frac{1}{3} \int \frac{dx}{x^2+1} - \frac{1}{3} \int \frac{dx}{x^2+2^2}$.$I = \frac{1}{3} 	an^{-1} x - \frac{1}{3} \cdot \frac{1}{2} 	an^{-1} \frac{x}{2} + c$.$I = \frac{1}{3} 	an^{-1} x - \frac{1}{6} 	an^{-1} \frac{x}{2} + c$.Comparing with $A 	an^{-1} x + B 	an^{-1} \frac{x}{2} + c$,we get $A = \frac{1}{3}$ and $B = -\frac{1}{6}$.

If $\int \frac{dx}{x^4+5x^2+4} = A \tan^{-1} x + B \tan^{-1} \frac{x}{2} + c$,where $c$ is a constant of integration,then:

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