evaluation of $\int\sqrt{\frac{e^{nx}}{e^{nx}+1}}dx:n\in \mathbb{Z^*}$

Question

Compute the indefinite integral

$$\displaystyle \int\sqrt{\frac{e^{nx}}{e^{nx}+1}}dx:n\in \mathbb{Z^*}$$

My Attempt:(The steps are shortened)

$$\displaystyle e^{nx}=\sinh^{2}(y)\Rightarrow dx=\frac{2}{n}\frac{\ \cosh(y)dy}{\sinh(y)}$$

$$\displaystyle \sqrt{\frac{e^{nx}}{e^{nx}+1}}dx=\frac{2}{n}dy$$

$$\displaystyle \int\sqrt{\frac{e^{nx}}{e^{nx}+1}}dx=\frac{2}{n}y+c=\frac{2}{n}\text{arcsinh}(\sqrt{e^{nx}})+c=\frac{2}{n}\ln(\sqrt{e^{nx}}+\sqrt{e^{nx}+1})+c$$

I am a new beginner so you should encourage me and not demoralize me. Thank you

Answer 1

This argument looks good to me. There doesn't appear to be any reason to restrict to integers here; the works just as well for any real value $n \neq 0$. Note that we can rewrite the antiderivative as $$x + \frac{2}{n} \log\left(1 + \sqrt{1 + e^{-n x}}\right) + C ,$$ which makes the asymptotic behavior as $x \to \infty$ clearer (for $n > 0$, anyway).

Alternatively,

• the substitution $$u = e^{nx / 2}, \qquad \,du = \frac{n}{2} e^{nx / 2}$$ transforms the integral to $$\frac{2}{n} \int \frac{du}{\sqrt{1 + u^2}} = \frac{2}{n} \operatorname{arsinh} u + C,$$ or
• the substitution $$e^{nx / 2} = \tan \theta, \qquad \frac{n}{2} e^{nx / 2} \,dx = \sec^2 \theta \,d\theta$$ transforms the integral to $$\frac{2}{n} \int \sec \theta \,d\theta = \frac{2}{n} \log \left\vert\sec\theta + \tan\theta\right\vert +C.$$

Answer 2

Set $y=e^{\frac{nx}{2}}$. Then the integral becomes

$$\int{\dfrac{y}{\sqrt{y^{2}+1}}}\dfrac{2}{n}\dfrac{dy}{y} =\dfrac{2}{n} \ln(y+\sqrt{y^{2}+1}).$$ Setting $y=e^{\frac{nx}{2}}$ we obtain

$$I=\dfrac{2}{n}\ln(\sqrt{e^{nx}+1}+e^{\frac{nx}{2}}).$$

Answer 3

Let $u=\sqrt{\frac{e^{nx}}{e^{nx}+1}}, $

$$I=\int u\cdot\frac{2}{n}\left( \frac{1}{u}+\frac{u}{1-u^2}\right)du=\frac{2}{n}\int \frac{1}{1-u^2}du=\frac{1}{n}\ln\left|\frac{1+u}{1-u}\right|+C$$

Put substitution back and simplify, $$I=\frac{1}{n}\ln\left|\frac{\sqrt{e^{nx}+1}+\sqrt{e^{nx}}}{\sqrt{e^{nx}+1}-\sqrt{e^{nx}}}\right|+C=\frac{2}{n}\ln\left(\sqrt{e^{nx}+1}+\sqrt{e^{nx}}\right)+C$$

Answer 4

$$ \begin{aligned} I &=\int \frac{\sqrt{e^{n x}}}{\sqrt{e^{n x}+1}} d x \\ &=2 \int \frac{\sqrt{e^{n x}}}{n e^{n x}} d \left(\sqrt{e^{n x}+1}\right) \\ &=\frac{2}{n} \int \frac{d\left(\sqrt{e^{n x}+1}\right)}{\sqrt{\left(\sqrt{e^{n x}+1}\right)^2-1}} \\ &=\frac{2}{n} \cosh ^{-1}\left(\sqrt{e^{n x}+1}\right)+C \quad \textrm{ via } \sqrt{e^{n x}+1}=\cosh \theta \end{aligned} $$