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Integral of d{x}:
Integral of ln(1+tgx)
Integral of 1/(2x+3)^4
Integral of sin(0,5x)
Integral of 1/shx
Identical expressions
ln(one +tgx)
ln(1 plus tgx)
ln(one plus tgx)
ln1+tgx
ln(1+tgx)dx
Similar expressions
ln(1-tgx)

Integral of ln(1+tgx) dx

The solution

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

Integral(log(1 + tan(x)), (x, 0, 1))

Detail solution

Use integration by parts:

$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$

Let $u{\left(x \right)} = \log{\left(\tan{\left(x \right)} + 1 \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$ .

Then $\operatorname{du}{\left(x \right)} = \frac{\tan^{2}{\left(x \right)} + 1}{\tan{\left(x \right)} + 1}$ .

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

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

The answer is:

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

The answer (Indefinite) [src]

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

$$x\,\log \left(\tan x+1\right)-{{x\,\log \left({{\tan ^2x+2\,\tan x+ 1}\over{2}}\right)-{\rm atan2}\left({{\tan x+1}\over{2}} , {{\tan x+ 1}\over{2}}\right)\,\log \left(\tan ^2x+1\right)-i\,{\it li}_{2}({{i \,\left(\left(i+1\right)\,\tan x-i+1\right)}\over{2}})+i\,{\it li}_{ 2}({{i\,\left(\left(i-1\right)\,\tan x-i-1\right)}\over{2}})}\over{2 }}$$

Simplify

The answer [src]

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0

$$-{{4\,\log \left({{\tan ^21+2\,\tan 1+1}\over{2}}\right)-\pi\,\log \left(\tan ^21+1\right)+4\,i\,{\it li}_{2}(-{{\left(i+1\right)\, \tan 1+i-1}\over{2}})-4\,i\,{\it li}_{2}({{\left(i-1\right)\,\tan 1+ i+1}\over{2}})}\over{8}}+\log \left(\tan 1+1\right)-{{i\,{\it li}_{2 }({{i+1}\over{2}})-i\,{\it li}_{2}(-{{i-1}\over{2}})}\over{2}}$$

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0

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

Simplify

Numerical answer [src]

0.446043170167152

0.446043170167152

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

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How to use it?

Identical expressions

Similar expressions

Integral of ln(1+tgx) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2