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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

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

Similar Questions

$\int {\frac{{p{x^{p + 2q - 1}} - q{x^{q - 1}}}}{{{x^{2p + 2q}} + 2{x^{p + q}} + 1}}dx} $ is equal to:

$\int \frac{dx}{(1 + x^2)\sqrt{p^2 + q^2(\tan^{-1}x)^2}} = $

If $\int \frac{dx}{e^x + 4e^{-x}} = f(x) + c$,then $f(x)$ is

If $f(x) = \int \frac{5x^8 + 7x^6}{(x^2 + 1 + 2x^7)^2} dx, x \geq 0$ and $f(0) = 0$,then the value of $f(1)$ is

$\int \frac{e^{2x}}{\sqrt[4]{e^x+1}} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module