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 resp

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

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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 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 integral formula $\int \frac{du}{\sqrt{u^2-a^2}} = \log |u + \sqrt{u^2-a^2}| + c$, we have $I = \frac{1}{\log 3} \log |z + \sqrt{z^2-1}| + c$.Finally, substituting $z = 3^x$ back, 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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