I tried to use binomial expansion for this problem, but it was in vain..I need the right approach. Can anyone guide me with the right idea?
$$\int_1^{\infty}\frac{\,dx}{e^{x+1}+e^{3-x}}$$
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Hint: Try the substitution $t = x-1$, and take $e^2$ common in the denominator. – stochasticboy321 Oct 17 '15 at 08:13
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Make the most obvious substitution. – A.S. Oct 17 '15 at 08:14
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\begin{align*} \int_1^{\infty}\frac{\,dx}{e^{x+1}+e^{3-x}} & = \int_1^{\infty}\frac{e^x}{e^{2x+1}+e^{3}} \, dx \end{align*} Let $e^{x}=t$, then \begin{align*} \int_1^{\infty}\frac{e^x}{e^{2x+1}+e^{3}} \, dx &= \frac{1}{e}\int_e^{\infty}\frac{1}{t^2+e^{2}} \, dx \end{align*} Now use $\arctan$ etc..
Anurag A
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