MHB Solve Exponential Integral: \int \frac{2^{x}\cdot 3^{x}}{9^{x}-4^{x}}dx

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The discussion focuses on solving the exponential integral \(\int \frac{2^{x}\cdot 3^{x}}{9^{x}-4^{x}}dx\). A user seeks hints to tackle this complex problem. A suggested approach involves rewriting the integral as \(\int \frac{1}{\frac{3^{x}}{2^x}-\frac{2^{x}}{3^x}}dx\). This transformation aims to simplify the expression for easier integration. The conversation highlights the challenge of the integral and the need for strategic manipulation to find a solution.
Yankel
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Hello

I am trying to solve this exponential integral, it's quite complicated. Any hints ?

\int \frac{2^{x}\cdot 3^{x}}{9^{x}-4^{x}}dxmany thanks
 
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\int \frac{ 1}{\frac{3^{x}}{2^x}-\frac{2^{x}}{3^x}}dx
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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