int frac{3^x}{sqrt{9^x-1}} dx = | vedclass

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Question

(A) Let $I = \int \frac{3^x}{\sqrt{9^x-1}} dx$. Substitute $3^x = t$. Then, differentiating both sides with respect to $x$, we get $3^x \log 3 dx = dt

Vedclass · Accepted Answer

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

Vedclass · Answer

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

$\int \frac{3^x}{\sqrt{9^x-1}} dx =$

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