Evaluate $\int \frac{e^{x}-1}{e^{x}+1} d x$
$\int \frac{e^{x}-1}{e^{x}+1} d x$
We can write above integrand as:
$\int\left(\frac{e^{x}}{e^{x}+1}-\frac{1}{e^{x}+1}\right) d x$
Considering integrand (A)
$A=\int \frac{e^{x}}{e^{x}+1} d x$
Put $e^{x}+1=t$
Differentiating w.r.t $x$ we get,
$e^{x} d x=d t$
Substituting values we get
$A=\int \frac{e^{x}}{e^{x}+1} d x=\int \frac{d t}{t} d x=\ln |t|+C$
Substituting the value of t we get,
$A=\ln \left|e^{x}+1\right|+C$
$\therefore A=\int \frac{e^{x}}{e^{x}+1} d x=\ln \left|e^{x}+1\right|+C \cdots(\mathrm{i})$
Considering integrand (B)
$B=\int \frac{1}{e^{x}+1} d x$
We can write above integral as
$\Rightarrow \int \frac{1+e^{x}-e^{x}}{e^{x}+1} d x$
(1) (2)
Considering first integral:
$\int \frac{1+e^{x}}{1+e^{x}} d x$
Since the numerator and denominator are exactly same, our integrand simplifies to 1 and integrand becomes:
$\Rightarrow \int \mathrm{dx}$
$\Rightarrow \mathrm{x}$
$\therefore \int \frac{1+e^{x}}{1+e^{x}} d x=x$
Let $u=1+e^{x}, d u=e^{x} d x$
Apply u - substitution:
$\int \frac{1}{\mathrm{u}}(-\mathrm{du})=-\ln |u|$
Replacing the value of u we get,
$\int \frac{-e^{x}}{\sigma^{x}+1} d x=-\ln \left|1+e^{x}\right|+C \cdots(4)$
From (3) and (4) we get,
$\Rightarrow \int \frac{1+e^{x}}{e^{x}+1} d x+\int \frac{-e^{x}}{e^{x}+1} d x=x-\ln \left|1+e^{x}\right|+C$
$\therefore \mathrm{B}=\int \frac{1}{e^{x}+1} d x=x-\ln \left|1+e^{x}\right|+C-(\mathrm{ii})$
From (i) and (ii) we get,
$\int \frac{e^{x}}{e^{x}+1} d x-\int \frac{1}{e^{x}+1} d x=\left(\ln \left|e^{x}+1\right|-\left(x-\ln \left|1+e^{x}\right|\right)\right)+C$
$=2 \ln \left|e^{x}+1\right|-x+C$
$\therefore \int \frac{e^{x}-1}{e^{x}+1} d x=2 \ln \left|e^{x}+1\right|-x+C$
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