# How to solve this integral?

### How to solve this integral?

$$\int\limits_{-1}^{1}\frac{1}{3^{x}+3^{-x}}dx$$
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### Re: How to solve this integral?

$$3^x= e^{ln(3^x)}= e^{x ln(3)}$$ and $$3^{-x}= e^{ln(3^{-x})}= e^{-x ln(3)}$$.
So $$\frac{1}{3^x+ 3^{-x}}=$$$$\frac{1}{e^{x ln(3)}+ e^{-x ln(3)}}=$$$$\frac{1}{2}\frac{2}{e^{x ln(3)}+ e^{-x ln(3)}}$$.

Now, cosh(x) is defined as $$\frac{e^x+ e^{-x}}{2}$$ so that is $$\frac{1}{2}\frac{1}{cosh(x ln(3))}$$. The integral is $$\frac{1}{2}\int_{-1}^1 \frac{dx}{cosh(xln(3))}$$

Letting $$u= x ln(3)$$, $$du= ln(3) dx$$, when x= 1, u= ln(3), and when x= -1, u= -ln(3) so that becomes $$\frac{1}{2 ln(3)}\int_{-ln(3)}^{ln(3)} \frac{du}{cosh(u)}$$.

That last integral is given at https://au.answers.yahoo.com/question/i ... 721AAbWgg9

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