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So i had to solve this integral:

$$\int_{0}^{1} \frac{a^x-a}{a^x+a} dx$$

$a\in\mathbb{R^+}$

So first i used substitution:

$t=a^x+a \implies a^x=t-a $

$ dx= \frac{dt}{\ln(a)(t-a)}$

Then with partial fractions i got this:

$$\frac{1}{\ln a} \left(2 \int_{1+a}^{2a}\frac{1}{t}dt-\int_{1+a}^{2a}\frac{1}{t-a}dt \right)$$

And with that i just used the common integral and have gotten this solution afterwards:

$$\frac{\ln(2a(1+a)^2)}{\ln a}$$

So i wonder if i did it right, at most the problem is with substitution but i think i did change bounds correctly, but any insight would be helpful. Thank you in advance.

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    $\begingroup$ if $a=1$? then integral gives 0, and your solution is not defined. maybe additionally $a\neq1$ $\endgroup$
    – serg_1
    Jun 21, 2017 at 9:05
  • $\begingroup$ Well, i may have made a mistake in the end when inserting bounds and calculating it, so i think$ a>0$ is enough $\endgroup$ Jun 21, 2017 at 9:28

4 Answers 4

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I think it is the correct substitution. We have

$$\int_0^1\frac{a^x-a}{a^x+1}dx=\frac{1}{\ln a} \left(2 \int_{1+a}^{2a}\frac{1}{t}dt-\int_{1+a}^{2a}\frac{1}{t-a}dt\right)$$

But the answer seems to be incorrect. Indeed,

\begin{align} &\;\frac{1}{\ln a} \left(2 \int_{1+a}^{2a}\frac{1}{t}dt-\int_{1+a}^{2a}\frac{1}{t-a}dt\right)\\ =&\;\frac{1}{\ln a}\Big[2\ln t-\ln (t-a)\Big]_{1+a}^{2a}\\ =&\;\frac{1}{\ln a}[2\ln2a-\ln a-2\ln(1+a)+\ln 1]\\ =&\;\frac{1}{\ln a}[2\ln2+\ln a-2\ln(1+a)]\\ =&\;\frac{\ln\left(\frac{4a}{(1+a)^2}\right)}{\ln a}\\ \end{align}

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May be, you could have started writing $$\int_{0}^{1} \frac{a^x-a}{a^x+a} dx=\int_{0}^{1} \frac{a^x+a-2a}{a^x+a} dx=\int_{0}^{1} dx-2a\int_{0}^{1} \frac{dx}{a^x+a}= 1-2a\int_{0}^{1} \frac{dx}{a^x+a}$$ Now, using your change of variable $$t=a^x+a \implies a^x=t-a\implies x=\frac{\log (t-a)}{\log (a)}\implies dx=\frac{dt}{(t-a) \log (a)}$$ makes (as you did) $$I=\int \frac{dx}{a^x+a}=\frac{1}{ \log (a)}\int \frac{dt}{t (t-a) }=\frac{1}{ a\log (a)}\int \left(\frac{1}{t-a}-\frac{1}{t} \right)dt=\frac{1}{ a\log (a)}\log\left(\frac{t-a}t\right)$$ and back to $x$ $$I=\frac{1}{ a\log (a)}\log\left(\frac{a^x}{a^x+a}\right)$$ Now, just use the bounds for $x$

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using a slightly different route: $$ \frac{a^x-a}{a^x+a} = \frac{a^{x-1}-1}{a^{x-1}+1} =\frac{e^{(x-1)\log a} -1}{e^{(x-1)\log a} +1} $$ set $u=(x-1)\log a$, then $$ I=\int_0^1 \frac{a^x-a}{a^x+a} dx = \frac1{\log a}\int_{-\log a}^0 \frac{e^u-1}{e^u+1}du = \frac1{\log a}\int_{-\log a}^0 \tanh \frac{u}{2} \quad du \\ $$ this gives $$ I = \frac2{\log a} \bigg[ \log \cosh \frac{u}{2}\bigg]_{-\log a}^0 \\ = \frac1{\log a} \bigg[ \log \cosh^2 \frac{u}{2}\bigg]_{-\log a}^0 \\ = \frac1{\log a} \bigg[ \log \frac{1+\cosh u}2 \bigg]_{-\log a}^0 \\ = \frac{\ln\left(\frac{4a}{(1+a)^2}\right)}{\ln a}\\ $$

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  • $\begingroup$ This is a nice change of variable ! Thanks for providing it and $+1$. Cheers. $\endgroup$ Jun 21, 2017 at 18:14
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hint: setting $$a^x=t+a$$ then we get$$x\ln(a)=t+a$$ and $$dx=\frac{1}{\ln(a)}dt$$

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