If int frac{dx}{(x^2+9) sqrt{x^2+16}} = frac{ | vedclass

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(A) Let $I = \int \frac{dx}{(x^2+9) \sqrt{x^2+16}}$.Put $x = 4 	an 	heta$, so $dx = 4 \sec^2 	heta \ d	heta$.Then $\sqrt{x^2+16} = \sqrt{16 	an^2

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$\frac{\sqrt{7}}{3}$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{(x^2+9) \sqrt{x^2+16}}$.Put $x = 4 	an 	heta$, so $dx = 4 \sec^2 	heta \ d	heta$.Then $\sqrt{x^2+16} = \sqrt{16 	an^2 	heta + 16} = 4 \sec 	heta$.Substituting these into the integral:$I = \int \frac{4 \sec^2 	heta \ d	heta}{(16 	an^2 	heta + 9)(4 \sec 	heta)} = \int \frac{\sec 	heta \ d	heta}{16 	an^2 	heta + 9} = \int \frac{\frac{1}{\cos 	heta} \ d	heta}{16 \frac{\sin^2 	heta}{\cos^2 	heta} + 9} = \int \frac{\cos 	heta \ d	heta}{16 \sin^2 	heta + 9 \cos^2 	heta}$.Since $\cos^2 	heta = 1 - \sin^2 	heta$, we have $16 \sin^2 	heta + 9(1 - \sin^2 	heta) = 7 \sin^2 	heta + 9$.So $I = \int \frac{\cos 	heta \ d	heta}{7 \sin^2 	heta + 9}$.Let $u = \sin 	heta$, then $du = \cos 	heta \ d	heta$.$I = \int \frac{du}{7u^2 + 9} = \frac{1}{7} \int \frac{du}{u^2 + (3/\sqrt{7})^2} = \frac{1}{7} \cdot \frac{\sqrt{7}}{3} \operatorname{Tan}^{-1} \left( \frac{u \sqrt{7}}{3} ight) + c = \frac{1}{3 \sqrt{7}} \operatorname{Tan}^{-1} \left( \frac{\sqrt{7} \sin 	heta}{3} ight) + c$.Since $x = 4 	an 	heta$, $\sin 	heta = \frac{x}{\sqrt{x^2+16}}$.Thus, $I = \frac{1}{3 \sqrt{7}} \operatorname{Tan}^{-1} \left( \frac{\sqrt{7}}{3} \cdot \frac{x}{\sqrt{x^2+16}} ight) + c$.Comparing this with the given expression, $K = \frac{\sqrt{7}}{3}$.

If $\int \frac{dx}{(x^2+9) \sqrt{x^2+16}} = \frac{1}{3 \sqrt{7}} \operatorname{Tan}^{-1} \left( K \frac{x}{\sqrt{16+x^2}} \right) + c$, then $K=$

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