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Integral of d{x}:
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Identical expressions
arctg2x-(three /(x^ four))arctg2x
arctg2x minus (3 divide by (x to the power of 4))arctg2x
arctg2x minus (three divide by (x to the power of four))arctg2x
arctg2x-(3/(x4))arctg2x
arctg2x-3/x4arctg2x
arctg2x-(3/(x⁴))arctg2x
arctg2x-3/x^4arctg2x
arctg2x-(3 divide by (x^4))arctg2x
arctg2x-(3/(x^4))arctg2xdx
Similar expressions
arctg2x+(3/(x^4))arctg2x

Solving integrals/ arctg2x-(3/(x^4))arctg2x

Integral of arctg2x-(3/(x^4))arctg2x dx

The solution

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  1                              
  /                              
 |                               
 |  /            3           \   
 |  |atan(2*x) - --*atan(2*x)| dx
 |  |             4          |   
 |  \            x           /   
 |                               
/                                
0

$$\int\limits_{0}^{1} \left(- \frac{3}{x^{4}} \operatorname{atan}{\left(2 x \right)} + \operatorname{atan}{\left(2 x \right)}\right)\, dx$$

Integral(atan(2*x) - 3/x^4*atan(2*x), (x, 0, 1))

Detail solution

Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{3}{x^{4}} \operatorname{atan}{\left(2 x \right)}\right)\, dx = - \int \frac{3 \operatorname{atan}{\left(2 x \right)}}{x^{4}}\, dx$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{3 \operatorname{atan}{\left(2 x \right)}}{x^{4}}\, dx = 3 \int \frac{\operatorname{atan}{\left(2 x \right)}}{x^{4}}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = \operatorname{atan}{\left(2 x \right)}$ and let $\operatorname{dv}{\left(x \right)} = \frac{1}{x^{4}}$ .
      
      Then $\operatorname{du}{\left(x \right)} = \frac{2}{4 x^{2} + 1}$ .
      
      To find $v{\left(x \right)}$ :
      1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
        
        $\int \frac{1}{x^{4}}\, dx = - \frac{1}{3 x^{3}}$
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{2}{3 x^{3} \left(4 x^{2} + 1\right)}\right)\, dx = - \frac{2 \int \frac{1}{x^{3} \left(4 x^{2} + 1\right)}\, dx}{3}$
      1. There are multiple ways to do this integral.
        
        Method #1
        
        Let $u = x^{2}$ .
        
        Then let $du = 2 x dx$ and substitute $du$ :
        
        $\int \frac{1}{8 u^{3} + 2 u^{2}}\, du$
        
        Rewrite the integrand:
        
        $\frac{1}{8 u^{3} + 2 u^{2}} = \frac{8}{4 u + 1} - \frac{2}{u} + \frac{1}{2 u^{2}}$
        
        Integrate term-by-term:
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{8}{4 u + 1}\, du = 8 \int \frac{1}{4 u + 1}\, du$
        
        Let $u = 4 u + 1$ .
        
        Then let $du = 4 du$ and substitute $\frac{du}{4}$ :
        
        $\int \frac{1}{4 u}\, du$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{4}$
        
        The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
        
        So, the result is: $\frac{\log{\left(u \right)}}{4}$
        
        Now substitute $u$ back in:
        
        $\frac{\log{\left(4 u + 1 \right)}}{4}$
        
        So, the result is: $2 \log{\left(4 u + 1 \right)}$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \left(- \frac{2}{u}\right)\, du = - 2 \int \frac{1}{u}\, du$
        
        The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
        
        So, the result is: $- 2 \log{\left(u \right)}$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{1}{2 u^{2}}\, du = \frac{\int \frac{1}{u^{2}}\, du}{2}$
        
        The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
        
        $\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$
        
        So, the result is: $- \frac{1}{2 u}$
        
        The result is: $- 2 \log{\left(u \right)} + 2 \log{\left(4 u + 1 \right)} - \frac{1}{2 u}$
        
        Now substitute $u$ back in:
        
        $- 2 \log{\left(x^{2} \right)} + 2 \log{\left(4 x^{2} + 1 \right)} - \frac{1}{2 x^{2}}$
        
        Method #2
        
        Rewrite the integrand:
        
        $\frac{1}{x^{3} \left(4 x^{2} + 1\right)} = \frac{16 x}{4 x^{2} + 1} - \frac{4}{x} + \frac{1}{x^{3}}$
        
        Integrate term-by-term:
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{16 x}{4 x^{2} + 1}\, dx = 16 \int \frac{x}{4 x^{2} + 1}\, dx$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{x}{4 x^{2} + 1}\, dx = \frac{\int \frac{8 x}{4 x^{2} + 1}\, dx}{8}$
        
        Let $u = 4 x^{2} + 1$ .
        
        Then let $du = 8 x dx$ and substitute $\frac{du}{8}$ :
        
        $\int \frac{1}{8 u}\, du$
        
        The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
        
        Now substitute $u$ back in:
        
        $\log{\left(4 x^{2} + 1 \right)}$
        
        So, the result is: $\frac{\log{\left(4 x^{2} + 1 \right)}}{8}$
        
        So, the result is: $2 \log{\left(4 x^{2} + 1 \right)}$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \left(- \frac{4}{x}\right)\, dx = - 4 \int \frac{1}{x}\, dx$
        
        The integral of $\frac{1}{x}$ is $\log{\left(x \right)}$ .
        
        So, the result is: $- 4 \log{\left(x \right)}$
        
        The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
        
        $\int \frac{1}{x^{3}}\, dx = - \frac{1}{2 x^{2}}$
        
        The result is: $- 4 \log{\left(x \right)} + 2 \log{\left(4 x^{2} + 1 \right)} - \frac{1}{2 x^{2}}$
        
        Method #3
        
        Rewrite the integrand:
        
        $\frac{1}{x^{3} \left(4 x^{2} + 1\right)} = \frac{1}{4 x^{5} + x^{3}}$
        
        Rewrite the integrand:
        
        $\frac{1}{4 x^{5} + x^{3}} = \frac{16 x}{4 x^{2} + 1} - \frac{4}{x} + \frac{1}{x^{3}}$
        
        Integrate term-by-term:
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{16 x}{4 x^{2} + 1}\, dx = 16 \int \frac{x}{4 x^{2} + 1}\, dx$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{x}{4 x^{2} + 1}\, dx = \frac{\int \frac{8 x}{4 x^{2} + 1}\, dx}{8}$
        
        Let $u = 4 x^{2} + 1$ .
        
        Then let $du = 8 x dx$ and substitute $\frac{du}{8}$ :
        
        $\int \frac{1}{8 u}\, du$
        
        The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
        
        Now substitute $u$ back in:
        
        $\log{\left(4 x^{2} + 1 \right)}$
        
        So, the result is: $\frac{\log{\left(4 x^{2} + 1 \right)}}{8}$
        
        So, the result is: $2 \log{\left(4 x^{2} + 1 \right)}$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \left(- \frac{4}{x}\right)\, dx = - 4 \int \frac{1}{x}\, dx$
        
        The integral of $\frac{1}{x}$ is $\log{\left(x \right)}$ .
        
        So, the result is: $- 4 \log{\left(x \right)}$
        
        The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
        
        $\int \frac{1}{x^{3}}\, dx = - \frac{1}{2 x^{2}}$
        
        The result is: $- 4 \log{\left(x \right)} + 2 \log{\left(4 x^{2} + 1 \right)} - \frac{1}{2 x^{2}}$
        
        Method #4
        
        Rewrite the integrand:
        
        $\frac{1}{x^{3} \left(4 x^{2} + 1\right)} = \frac{1}{4 x^{5} + x^{3}}$
        
        Rewrite the integrand:
        
        $\frac{1}{4 x^{5} + x^{3}} = \frac{16 x}{4 x^{2} + 1} - \frac{4}{x} + \frac{1}{x^{3}}$
        
        Integrate term-by-term:
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{16 x}{4 x^{2} + 1}\, dx = 16 \int \frac{x}{4 x^{2} + 1}\, dx$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{x}{4 x^{2} + 1}\, dx = \frac{\int \frac{8 x}{4 x^{2} + 1}\, dx}{8}$
        
        Let $u = 4 x^{2} + 1$ .
        
        Then let $du = 8 x dx$ and substitute $\frac{du}{8}$ :
        
        $\int \frac{1}{8 u}\, du$
        
        The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
        
        Now substitute $u$ back in:
        
        $\log{\left(4 x^{2} + 1 \right)}$
        
        So, the result is: $\frac{\log{\left(4 x^{2} + 1 \right)}}{8}$
        
        So, the result is: $2 \log{\left(4 x^{2} + 1 \right)}$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \left(- \frac{4}{x}\right)\, dx = - 4 \int \frac{1}{x}\, dx$
        
        The integral of $\frac{1}{x}$ is $\log{\left(x \right)}$ .
        
        So, the result is: $- 4 \log{\left(x \right)}$
        
        The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
        
        $\int \frac{1}{x^{3}}\, dx = - \frac{1}{2 x^{2}}$
        
        The result is: $- 4 \log{\left(x \right)} + 2 \log{\left(4 x^{2} + 1 \right)} - \frac{1}{2 x^{2}}$
      So, the result is: $\frac{4 \log{\left(x^{2} \right)}}{3} - \frac{4 \log{\left(4 x^{2} + 1 \right)}}{3} + \frac{1}{3 x^{2}}$
    So, the result is: $- 4 \log{\left(x^{2} \right)} + 4 \log{\left(4 x^{2} + 1 \right)} - \frac{1}{x^{2}} - \frac{\operatorname{atan}{\left(2 x \right)}}{x^{3}}$
  So, the result is: $4 \log{\left(x^{2} \right)} - 4 \log{\left(4 x^{2} + 1 \right)} + \frac{1}{x^{2}} + \frac{\operatorname{atan}{\left(2 x \right)}}{x^{3}}$
1. There are multiple ways to do this integral.
  Method #1
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{\operatorname{atan}{\left(u \right)}}{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \operatorname{atan}{\left(u \right)}\, du = \frac{\int \operatorname{atan}{\left(u \right)}\, du}{2}$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = \operatorname{atan}{\left(u \right)}$ and let $\operatorname{dv}{\left(u \right)} = 1$ .
      
      Then $\operatorname{du}{\left(u \right)} = \frac{1}{u^{2} + 1}$ .
      
      To find $v{\left(u \right)}$ :
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u}{u^{2} + 1}\, du = \frac{\int \frac{2 u}{u^{2} + 1}\, du}{2}$
      
      Let $u = u^{2} + 1$ .
      
      Then let $du = 2 u du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{2 u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(u^{2} + 1 \right)}$
      
      So, the result is: $\frac{\log{\left(u^{2} + 1 \right)}}{2}$
      
      So, the result is: $\frac{u \operatorname{atan}{\left(u \right)}}{2} - \frac{\log{\left(u^{2} + 1 \right)}}{4}$
    Now substitute $u$ back in:
    
    $x \operatorname{atan}{\left(2 x \right)} - \frac{\log{\left(4 x^{2} + 1 \right)}}{4}$
  Method #2
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = \operatorname{atan}{\left(2 x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$ .
    
    Then $\operatorname{du}{\left(x \right)} = \frac{2}{4 x^{2} + 1}$ .
    
    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.
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{2 x}{4 x^{2} + 1}\, dx = 2 \int \frac{x}{4 x^{2} + 1}\, dx$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{x}{4 x^{2} + 1}\, dx = \frac{\int \frac{8 x}{4 x^{2} + 1}\, dx}{8}$
      
      Let $u = 4 x^{2} + 1$ .
      
      Then let $du = 8 x dx$ and substitute $\frac{du}{8}$ :
      
      $\int \frac{1}{8 u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(4 x^{2} + 1 \right)}$
      
      So, the result is: $\frac{\log{\left(4 x^{2} + 1 \right)}}{8}$
    So, the result is: $\frac{\log{\left(4 x^{2} + 1 \right)}}{4}$
The result is: $x \operatorname{atan}{\left(2 x \right)} + 4 \log{\left(x^{2} \right)} - \frac{17 \log{\left(4 x^{2} + 1 \right)}}{4} + \frac{1}{x^{2}} + \frac{\operatorname{atan}{\left(2 x \right)}}{x^{3}}$
Add the constant of integration:

$x \operatorname{atan}{\left(2 x \right)} + 4 \log{\left(x^{2} \right)} - \frac{17 \log{\left(4 x^{2} + 1 \right)}}{4} + \frac{1}{x^{2}} + \frac{\operatorname{atan}{\left(2 x \right)}}{x^{3}}+ \mathrm{constant}$

The answer is:

$x \operatorname{atan}{\left(2 x \right)} + 4 \log{\left(x^{2} \right)} - \frac{17 \log{\left(4 x^{2} + 1 \right)}}{4} + \frac{1}{x^{2}} + \frac{\operatorname{atan}{\left(2 x \right)}}{x^{3}}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                               
 |                                                            /       2\                          
 | /            3           \          1         / 2\   17*log\1 + 4*x /                 atan(2*x)
 | |atan(2*x) - --*atan(2*x)| dx = C + -- + 4*log\x / - ---------------- + x*atan(2*x) + ---------
 | |             4          |           2                      4                              3   
 | \            x           /          x                                                     x    
 |                                                                                                
/

$$\int \left(- \frac{3}{x^{4}} \operatorname{atan}{\left(2 x \right)} + \operatorname{atan}{\left(2 x \right)}\right)\, dx = C + x \operatorname{atan}{\left(2 x \right)} + 4 \log{\left(x^{2} \right)} - \frac{17 \log{\left(4 x^{2} + 1 \right)}}{4} + \frac{1}{x^{2}} + \frac{\operatorname{atan}{\left(2 x \right)}}{x^{3}}$$

Simplify

The answer [src]

-oo

$$-\infty$$

-oo

$$-\infty$$

-oo

Numerical answer [src]

-5.49219022742095e+38

-5.49219022742095e+38

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

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Integral of arctg2x-(3/(x^4))arctg2x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4

Method #1

Method #2