int frac{3^x , dx}{sqrt{9^x-1}} is equal to | vedclass

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(A) Let $I = \int \frac{3^x \, dx}{\sqrt{9^x-1}} = \int \frac{3^x \, dx}{\sqrt{(3^x)^2-1}}$.Substitute $3^x = z$. Then,differentiating both sides with

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$\frac{1}{\log 3} \log \left|3^x+\sqrt{9^x-1}\right|+c$

Vedclass · Answer

(A) Let $I = \int \frac{3^x \, dx}{\sqrt{9^x-1}} = \int \frac{3^x \, dx}{\sqrt{(3^x)^2-1}}$.Substitute $3^x = z$. Then,differentiating both sides with respect to $x$,we get $3^x \log 3 \, dx = dz$,which implies $3^x \, dx = \frac{dz}{\log 3}$.Substituting these into the integral,we get $I = \frac{1}{\log 3} \int \frac{dz}{\sqrt{z^2-1}}$.Using the standard integration formula $\int \frac{dt}{\sqrt{t^2-a^2}} = \log |t + \sqrt{t^2-a^2}| + c$,we obtain:$I = \frac{1}{\log 3} \log |z + \sqrt{z^2-1}| + c$.Substituting $z = 3^x$ back into the expression,we get:$I = \frac{1}{\log 3} \log |3^x + \sqrt{9^x-1}| + c$.

$\int \frac{3^x \, dx}{\sqrt{9^x-1}}$ is equal to

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