int frac{dx}{4x^2 + 9} = | vedclass

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Question

(C) We need to evaluate the integral $I = \int \frac{dx}{4x^2 + 9}$.Factor out $4$ from the denominator:$I = \int \frac{dx}{4(x^2 + \frac{9}{4})} = \f

Vedclass · Accepted Answer

$\frac{1}{6} 	an^{-1} \left( \frac{2x}{3} ight) + c$

Vedclass · Answer

(C) We need to evaluate the integral $I = \int \frac{dx}{4x^2 + 9}$.Factor out $4$ from the denominator:$I = \int \frac{dx}{4(x^2 + \frac{9}{4})} = \frac{1}{4} \int \frac{dx}{x^2 + (\frac{3}{2})^2}$.Using the standard integral formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$,where $a = \frac{3}{2}$:$I = \frac{1}{4} \cdot \frac{1}{3/2} 	an^{-1}(\frac{x}{3/2}) + c$.$I = \frac{1}{4} \cdot \frac{2}{3} 	an^{-1}(\frac{2x}{3}) + c$.$I = \frac{1}{6} 	an^{-1}(\frac{2x}{3}) + c$.

$\int \frac{dx}{4x^2 + 9} = $

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