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I solved this integral: $$I=\int\limits_{0}^{4} \frac{dx}{4+2^x} $$ In a method similar to the one used in An interesting integral $I = \int\limits_{-1}^{1} \arctan(e^x)dx $. This integral comes from the following article http://www.maa.org/sites/default/files/pdf/mathdl/CMJ/Nelsen39-41.pdf.

The substitution $x = u+2$ gives: $$I = \int\limits_{-2}^{2} \frac{du}{4+2^{u+2}}=\int\limits_{-2}^{2}f(u)du $$ Now if we look at $g(u) = f(u) - 0.125 = \frac{1}{4+2^{u+2}} - 0.125 = \frac{1}{4} \left[ \frac{1}{1+2^{u}} -0.5 \right]$ we can see that \begin{eqnarray} g(u)=\frac{1}{4} \left[ \frac{1}{1+2^{u}} -0.5 \right] &=& \frac{1}{4} \left[ \frac{2^{-u}}{2^{-u}+1} - \underbrace{\left(\frac{2^{-u}+1}{2^{-u}+1} - 0.5\right) }_{=0.5} \right]\\ &=& \frac{1}{4} \left[ \frac{-1}{2^{-u}+1} +0.5 \right] \\ &=&\frac{-1}{4+2^{-u+2}}+ 0.125\\ &=& -g(-u). \end{eqnarray} This means that $g(u)$ is an odd function. We see that $f(u) = g(u)+0.125$ and thus we have that $$I=\int\limits_{-2}^{2}f(u)du =\int\limits_{-2}^{2}g(u)du +\int\limits_{-2}^{2}0.125du = 0.5 .$$ Here I used the fact that $\int\limits_{-2}^{2}g(u)du=0$ because $g(u)$ is even.

So this method is pretty nice and straightforward but what are other ways to solve this integral? I feel that there should be other ways to solve this integral over an other integration interval for instance. If we were integrating from $0$ to $5$ in the original integral for instance this trick would not have worked.

Wolfram says that $$\int \frac{dx}{4+2^x} dx=\frac{x}{4}-\frac{\log(4+2^x)}{4 \log(2)}$$ How would we show this? Thanks in advance!

EDIT The reason I asked this question and the linked one is that this method provided such as 'easy' way to solve the integral if you know about the symmetry of the integrand. I have another really interesting one: $\int\limits_{0}^{2\pi} \frac{1}{1+e^{\sin(x)}} dx =\pi$ which is easily shown once you see the symmetry. This is another example of an integral that seems very hard to me without this trick. Anyone have any ideas of solving it using standard methods? Thanks!

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Slugger
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2 Answers2

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Note that $$\int \frac{4+2^x}{4+2^x}\,dx=\int\frac{4}{4+2^x}\,dx+\int \frac{2^x}{4+2^x}\,dx.$$ For the last integral, let $u=4+2^x$.

André Nicolas
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    Thanks for the quick answer, this is a nice method! I also added another in the edit that you might like to try. Maybe I should post that one as a separate question? – Slugger Dec 16 '13 at 22:07
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Set $4+2^x = t$. We then have $2^x \ln(2) dx = dt$. Hence, $$dx = \dfrac{dt}{\ln(2)(t-4)}$$ Hence, $$\dfrac{dx}{4+2^x} = \dfrac{dt}{\ln(2) t (t-4)}$$ Trust you can finish off from here.

  • Ah thanks for your answer, I guess it wasn't as hard as I originally thought. Thanks for your answer! I also added another in the edit that you might like to try. Maybe I should post that one as a separate question? – Slugger Dec 16 '13 at 22:06