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Integral of d{x}:
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Identical expressions
(one +tg(x))/(one -tg(x))
(1 plus tg(x)) divide by (1 minus tg(x))
(one plus tg(x)) divide by (one minus tg(x))
1+tgx/1-tgx
(1+tg(x)) divide by (1-tg(x))
(1+tg(x))/(1-tg(x))dx
Similar expressions
((1+tgx)/(1-tgx))*dx
(1-tg(x))/(1-tg(x))
(1+tg(x))/(1+tg(x))

Solving integrals/ (1+tg(x))/(1-tg(x))

Integral of (1+tg(x))/(1-tg(x)) dx

The solution

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  1              
  /              
 |               
 |  1 + tan(x)   
 |  ---------- dx
 |  1 - tan(x)   
 |               
/                
0

$$\int\limits_{0}^{1} \frac{\tan{\left(x \right)} + 1}{- \tan{\left(x \right)} + 1}\, dx$$

Integral((1 + tan(x))/(1 - tan(x)), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{\tan{\left(x \right)} + 1}{1 - \tan{\left(x \right)}} = \frac{- \tan{\left(x \right)} - 1}{\tan{\left(x \right)} - 1}$
2. Rewrite the integrand:
  
  $\frac{- \tan{\left(x \right)} - 1}{\tan{\left(x \right)} - 1} = - \frac{\tan{\left(x \right)}}{\tan{\left(x \right)} - 1} - \frac{1}{\tan{\left(x \right)} - 1}$
3. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\tan{\left(x \right)}}{\tan{\left(x \right)} - 1}\right)\, dx = - \int \frac{\tan{\left(x \right)}}{\tan{\left(x \right)} - 1}\, dx$
    1. Don't know the steps in finding this integral.
      
      But the integral is
      
      $\frac{x}{2} + \frac{\log{\left(\tan{\left(x \right)} - 1 \right)}}{2} - \frac{\log{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{4}$
    So, the result is: $- \frac{x}{2} - \frac{\log{\left(\tan{\left(x \right)} - 1 \right)}}{2} + \frac{\log{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{\tan{\left(x \right)} - 1}\right)\, dx = - \int \frac{1}{\tan{\left(x \right)} - 1}\, dx$
    1. Don't know the steps in finding this integral.
      
      But the integral is
      
      $- \frac{x}{2} + \frac{\log{\left(\tan{\left(x \right)} - 1 \right)}}{2} - \frac{\log{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{4}$
    So, the result is: $\frac{x}{2} - \frac{\log{\left(\tan{\left(x \right)} - 1 \right)}}{2} + \frac{\log{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{4}$
  The result is: $- \log{\left(\tan{\left(x \right)} - 1 \right)} + \frac{\log{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{2}$
Method #2
1. Rewrite the integrand:
  
  $\frac{\tan{\left(x \right)} + 1}{1 - \tan{\left(x \right)}} = \frac{\tan{\left(x \right)}}{1 - \tan{\left(x \right)}} + \frac{1}{1 - \tan{\left(x \right)}}$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\frac{\tan{\left(x \right)}}{1 - \tan{\left(x \right)}} = - \frac{\tan{\left(x \right)}}{\tan{\left(x \right)} - 1}$
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\tan{\left(x \right)}}{\tan{\left(x \right)} - 1}\right)\, dx = - \int \frac{\tan{\left(x \right)}}{\tan{\left(x \right)} - 1}\, dx$
    1. Don't know the steps in finding this integral.
      
      But the integral is
      
      $\frac{x}{2} + \frac{\log{\left(\tan{\left(x \right)} - 1 \right)}}{2} - \frac{\log{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{4}$
    So, the result is: $- \frac{x}{2} - \frac{\log{\left(\tan{\left(x \right)} - 1 \right)}}{2} + \frac{\log{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{4}$
  1. Rewrite the integrand:
    
    $\frac{1}{1 - \tan{\left(x \right)}} = - \frac{1}{\tan{\left(x \right)} - 1}$
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{\tan{\left(x \right)} - 1}\right)\, dx = - \int \frac{1}{\tan{\left(x \right)} - 1}\, dx$
    1. Don't know the steps in finding this integral.
      
      But the integral is
      
      $- \frac{x}{2} + \frac{\log{\left(\tan{\left(x \right)} - 1 \right)}}{2} - \frac{\log{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{4}$
    So, the result is: $\frac{x}{2} - \frac{\log{\left(\tan{\left(x \right)} - 1 \right)}}{2} + \frac{\log{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{4}$
  The result is: $- \log{\left(\tan{\left(x \right)} - 1 \right)} + \frac{\log{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{2}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

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

$${{\log \left(\tan ^2x+1\right)}\over{2}}-\log \left(\tan x-1\right)$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

nan

$${{\log \left(\tan ^21+1\right)}\over{2}}-\log \left(\tan 1-1\right)$$

nan

$$\text{NaN}$$

Simplify

Numerical answer [src]

-1.5989832649916

-1.5989832649916

The graph

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

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Identical expressions

Similar expressions

Integral of (1+tg(x))/(1-tg(x)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2