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

If $\int \frac{e^x}{\sqrt{e^{2x}+4e^x+13}} dx = \log \left|e^x+2+\sqrt{e^{2x}+4e^x+13}\right|+c$,(where $c$ is the constant of integration),then the value of $a$ in the expression $\log \left|e^{ax}+2+\sqrt{e^{2x}+4e^x+13}\right|+c$ is equal to

$\int \sin \sqrt{x} \,d x=\ldots+C$ (where $C$ is a constant of integration.)

Evaluate the integral: $\int \frac{\cos ^4 x}{\left(\sin ^2 x+\sin ^{-3} x \cos ^5 x\right)^3} d x$

$\int \frac{d x}{\sqrt{\sin ^3 x \cos (x-\alpha)}}=$

Integrate the function $\frac{e^{2x}-1}{e^{2x}+1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module