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tanh(x)^2
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tanh(x)^2dx

Solving integrals/ tanh(x)^2

Integral of tanh(x)^2 dx

The solution

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  1            
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 |      2      
 |  tanh (x) dx
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0

$$\int\limits_{0}^{1} \tanh^{2}{\left(x \right)}\, dx$$

Integral(tanh(x)^2, (x, 0, 1))

Detail solution

Let $u = \tanh{\left(x \right)}$ .

Then let $du = \left(1 - \tanh^{2}{\left(x \right)}\right) dx$ and substitute $- du$ :

$\int \frac{u^{2}}{u^{2} - 1}\, du$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{u^{2}}{u^{2} - 1}\right)\, du = - \int \frac{u^{2}}{u^{2} - 1}\, du$
  1. Rewrite the integrand:
    
    $\frac{u^{2}}{u^{2} - 1} = 1 - \frac{1}{2 \left(u + 1\right)} + \frac{1}{2 \left(u - 1\right)}$
  2. Integrate term-by-term:
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{2 \left(u + 1\right)}\right)\, du = - \frac{\int \frac{1}{u + 1}\, du}{2}$
      1. Let $u = u + 1$ .
        
        Then let $du = du$ and substitute $du$ :
        
        $\int \frac{1}{u}\, du$
        
        The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
        
        Now substitute $u$ back in:
        
        $\log{\left(u + 1 \right)}$
      So, the result is: $- \frac{\log{\left(u + 1 \right)}}{2}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{2 \left(u - 1\right)}\, du = \frac{\int \frac{1}{u - 1}\, du}{2}$
      1. Let $u = u - 1$ .
        
        Then let $du = du$ and substitute $du$ :
        
        $\int \frac{1}{u}\, du$
        
        The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
        
        Now substitute $u$ back in:
        
        $\log{\left(u - 1 \right)}$
      So, the result is: $\frac{\log{\left(u - 1 \right)}}{2}$
    The result is: $u + \frac{\log{\left(u - 1 \right)}}{2} - \frac{\log{\left(u + 1 \right)}}{2}$
  So, the result is: $- u - \frac{\log{\left(u - 1 \right)}}{2} + \frac{\log{\left(u + 1 \right)}}{2}$
Now substitute $u$ back in:

$- \frac{\log{\left(\tanh{\left(x \right)} - 1 \right)}}{2} + \frac{\log{\left(\tanh{\left(x \right)} + 1 \right)}}{2} - \tanh{\left(x \right)}$
Add the constant of integration:

$- \frac{\log{\left(\tanh{\left(x \right)} - 1 \right)}}{2} + \frac{\log{\left(\tanh{\left(x \right)} + 1 \right)}}{2} - \tanh{\left(x \right)}+ \mathrm{constant}$

The answer is:

$- \frac{\log{\left(\tanh{\left(x \right)} - 1 \right)}}{2} + \frac{\log{\left(\tanh{\left(x \right)} + 1 \right)}}{2} - \tanh{\left(x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                
 |                                                                 
 |     2             log(1 + tanh(x))             log(-1 + tanh(x))
 | tanh (x) dx = C + ---------------- - tanh(x) - -----------------
 |                          2                             2        
/

$$x-{{2}\over{e^ {- 2\,x }+1}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

1 - tanh(1)

$$1-{{e^2-1}\over{e^2+1}}$$

1 - tanh(1)

$$1 - \tanh{\left(1 \right)}$$

Упростить

Numerical answer [src]

0.238405844044235

0.238405844044235

The graph

Use the examples entering the upper and lower limits of integration.

Other calculators

How to use it?

Identical expressions

Integral of tanh(x)^2 dx

Limits of integration:

The graph:

Piecewise:

The solution