How do you find the antiderivative of ${\frac{e^{x}}{(e^{2 x}) - 1}}$ ${(e^x)/ [(e^(2x)) - 1]}$ ?

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How do you find the antiderivative of ${\frac{e^{x}}{(e^{2 x}) - 1}}$ ${(e^x)/ [(e^(2x)) - 1]}$ ?

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Answer 1 · 2016-10-22T18:15:06

$ln (\frac{\sqrt{e^{2 x} - 1}}{e^{x} + 1}) + C$ $ln(sqrt(e^(2x)-1)/(e^x+1))+C$

Explanation:

$I = \int \frac{e^{x}}{e^{2 x} - 1} d x$ $I=inte^x/(e^(2x)-1)dx$

Let $u = e^{x}$ $u=e^x$ so that $d u = e^{x} d x$ $du=e^xdx$ :

$I = \int \frac{e^{x}}{{(e^{x})}^{2} - 1} d x = \int \frac{1}{u^{2} - 1} d u$ $I=inte^x/((e^x)^2-1)dx=int1/(u^2-1)du$

Now, we can ley $u = sec θ$ $u=sectheta$ . This implies that $d u = sec θ tan θ d θ$ $du=secthetatanthetad theta$ .

Plugging these in:

$I = \int \frac{sec θ tan θ}{{sec}^{2} θ - 1} d θ$ $I=int(secthetatantheta)/(sec^2theta-1)d theta$

Note that $1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2theta=sec^2theta$ , so ${sec}^{2} θ - 1 = {tan}^{2} θ$ $sec^2theta-1=tan^2theta$ :

$I = \int \frac{sec θ tan θ}{{tan}^{2} θ} d θ$ $I=int(secthetatantheta)/tan^2thetad theta$

$I = \int \frac{sec θ}{tan θ} d θ$ $I=intsectheta/tanthetad theta$

$I = \int \frac{1}{cos θ} (\frac{cos θ}{sin θ}) d θ$ $I=int1/costheta(costheta/sintheta)d theta$

$I = \int csc θ$ $I=intcsctheta$

This is a fairly common integral:

$I = - ln (| csc θ + cot θ |)$ $I=-ln(abs(csctheta+cottheta))$

We need to rewrite this using $u = sec θ$ $u=sectheta$ . This means we have a right triangle where $u$ $u$ is the hypotenuse, $1$ $1$ is the side adjacent to $θ$ $theta$ , and $\sqrt{u^{2} - 1}$ $sqrt(u^2-1)$ is the side opposite $θ$ $theta$ .

Thus $csc θ = \frac{u}{\sqrt{u^{2} - 1}}$ $csctheta=u/sqrt(u^2-1)$ and $cot θ = \frac{1}{\sqrt{u^{2} - 1}}$ $cottheta=1/sqrt(u^2-1)$ .

So:

$I = - ln (∣ ∣ ∣ \frac{u}{\sqrt{u^{2} - 1}} + \frac{1}{\sqrt{u^{2} - 1}} ∣ ∣ ∣) + C$ $I=-ln(abs(u/sqrt(u^2-1)+1/sqrt(u^2-1)))+C$

$I = - ln (∣ ∣ ∣ \frac{u + 1}{\sqrt{u^{2} - 1}} ∣ ∣ ∣) + C$ $I=-ln(abs((u+1)/sqrt(u^2-1)))+C$

Bringing in the negative $1$ $1$ as an exponent through the rule $B log (A) = log (A^{B})$ $Blog(A)=log(A^B)$ :

$I = ln (∣ ∣ ∣ \frac{\sqrt{u^{2} - 1}}{u + 1} ∣ ∣ ∣) + C$ $I=ln(abs(sqrt(u^2-1)/(u+1)))+C$

Through $u = e^{x}$ $u=e^x$ :

$I = ln (∣ ∣ ∣ \frac{\sqrt{e^{2 x} - 1}}{e^{x} + 1} ∣ ∣ ∣) + C$ $I=ln(abs(sqrt(e^(2x)-1)/(e^x+1)))+C$

Notice that $\sqrt{e^{2 x} - 1} > 0$ $sqrt(e^(2x)-1)>0$ and $e^{x} + 1 > 0$ $e^x+1>0$ for all values of $x$ $x$ , so the absolute value bars can be removed:

$I = ln (\frac{\sqrt{e^{2 x} - 1}}{e^{x} + 1}) + C$ $I=ln(sqrt(e^(2x)-1)/(e^x+1))+C$

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How do you find the antiderivative of ${\frac{e^{x}}{(e^{2 x}) - 1}}$ ${(e^x)/ [(e^(2x)) - 1]}$ ?

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