int frac{1}{16 x^{2}+9} d x is equal to | vedclass

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(C) We know that $\int \frac{1}{x^{2}+a^{2}} d x = \frac{1}{a} 	an ^{-1}\left(\frac{x}{a}ight)+c$. Given integral is $I = \int \frac{1}{16 x^{2}+9}

Vedclass · Accepted Answer

$\frac{1}{12} 	an ^{-1}\left(\frac{4 x}{3}ight)+c$

Vedclass · Answer

(C) We know that $\int \frac{1}{x^{2}+a^{2}} d x = \frac{1}{a} 	an ^{-1}\left(\frac{x}{a}ight)+c$. Given integral is $I = \int \frac{1}{16 x^{2}+9} d x$. Taking $16$ as a common factor from the denominator: $I = \frac{1}{16} \int \frac{1}{x^{2}+\frac{9}{16}} d x = \frac{1}{16} \int \frac{1}{x^{2}+\left(\frac{3}{4}ight)^{2}} d x$. Using the formula with $a = \frac{3}{4}$: $I = \frac{1}{16} 	imes \frac{1}{3/4} 	an ^{-1}\left(\frac{x}{3/4}ight)+c$. $I = \frac{1}{16} 	imes \frac{4}{3} 	an ^{-1}\left(\frac{4 x}{3}ight)+c$. $I = \frac{1}{12} 	an ^{-1}\left(\frac{4 x}{3}ight)+c$.

$\int \frac{1}{16 x^{2}+9} d x$ is equal to

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