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

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Question

(A) Let $I = \int \frac{3^x}{\sqrt{9^x - 1}} \, dx$. We can rewrite the integral as $I = \int \frac{3^x}{\sqrt{(3^x)^2 - 1}} \, dx$. Let $3^x = t$. Th

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{3^x}{\sqrt{9^x - 1}} \, dx$. We can rewrite the integral as $I = \int \frac{3^x}{\sqrt{(3^x)^2 - 1}} \, dx$. Let $3^x = t$. 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 $\int \frac{dt}{\sqrt{t^2 - a^2}} = \log |t + \sqrt{t^2 - a^2}| + c$,we obtain $I = \frac{1}{\log 3} \log |t + \sqrt{t^2 - 1}| + c$. Finally,substituting $t = 3^x$ back,we get $I = \frac{1}{\log 3} \log |3^x + \sqrt{9^x - 1}| + c$.

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

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