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

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

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{3^x \, dx}{\sqrt{9^x-1}} = \int \frac{3^x \, dx}{\sqrt{(3^x)^2-1}}$.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 integration formula $\int \frac{dt}{\sqrt{t^2-a^2}} = \log |t + \sqrt{t^2-a^2}| + c$,we obtain:$I = \frac{1}{\log 3} \log |z + \sqrt{z^2-1}| + c$.Substituting $z = 3^x$ back into the expression,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

Similar Questions

$\int \frac{\sin x \cos x}{a \cos^2 x + b \sin^2 x} dx = $

$\int \frac{(3 x-2) \tan \left(\sqrt{9 x^2-12 x+1}\right)}{\sqrt{9 x^2-12 x+1}} d x=$

The value of the integral $\int \frac{d x}{\left(e^x+e^{-x}\right)^2}$ is

If $\int \frac{1}{(1 + x)\sqrt{x}} \, dx = f(x) + A$,where $A$ is any arbitrary constant,then the function $f(x)$ is

$\int \sqrt{\frac{x}{a^3 - x^3}} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module