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Integral of d{x}:
Integral of 1/(x^2+3)
Integral of x^-3
Integral of a^x
Integral of dy/y^2
Identical expressions
(x^ two - four x+ five)/((x+ two)(x^ two +4))
(x squared minus 4x plus 5) divide by ((x plus 2)(x squared plus 4))
(x to the power of two minus four x plus five) divide by ((x plus two)(x to the power of two plus 4))
(x2-4x+5)/((x+2)(x2+4))
x2-4x+5/x+2x2+4
(x²-4x+5)/((x+2)(x²+4))
(x to the power of 2-4x+5)/((x+2)(x to the power of 2+4))
x^2-4x+5/x+2x^2+4
(x^2-4x+5) divide by ((x+2)(x^2+4))
(x^2-4x+5)/((x+2)(x^2+4))dx
Similar expressions
(x^2-4x+5)/((x-2)(x^2+4))
(x^2-4x+5)/((x+2)(x^2-4))
(x^2+4x+5)/((x+2)(x^2+4))
(x^2-4x-5)/((x+2)(x^2+4))

Solving integrals/ (x^2-4x+5)/((x+2)(x^2+4))

Integral of (x^2-4x+5)/((x+2)(x^2+4)) dx

The solution

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  1                    
  /                    
 |                     
 |     2               
 |    x  - 4*x + 5     
 |  ---------------- dx
 |          / 2    \   
 |  (x + 2)*\x  + 4/   
 |                     
/                      
0

$$\int\limits_{0}^{1} \frac{x^{2} - 4 x + 5}{\left(x + 2\right) \left(x^{2} + 4\right)}\, dx$$

Integral((x^2 - 4*x + 5)/(((x + 2)*(x^2 + 4))), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{x^{2} - 4 x + 5}{\left(x + 2\right) \left(x^{2} + 4\right)} = - \frac{9 x + 14}{8 \left(x^{2} + 4\right)} + \frac{17}{8 \left(x + 2\right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{9 x + 14}{8 \left(x^{2} + 4\right)}\right)\, dx = - \frac{\int \frac{9 x + 14}{x^{2} + 4}\, dx}{8}$
    1. Rewrite the integrand:
      
      $\frac{9 x + 14}{x^{2} + 4} = \frac{9 x}{x^{2} + 4} + \frac{14}{x^{2} + 4}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{9 x}{x^{2} + 4}\, dx = \frac{9 \int \frac{2 x}{x^{2} + 4}\, dx}{2}$
      
      Let $u = x^{2} + 4$ .
      
      Then let $du = 2 x dx$ and substitute $\frac{9 du}{2}$ :
      
      $\int \frac{9}{2 u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x^{2} + 4 \right)}$
      
      So, the result is: $\frac{9 \log{\left(x^{2} + 4 \right)}}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{14}{x^{2} + 4}\, dx = 14 \int \frac{1}{x^{2} + 4}\, dx$
      
      The integral of $\frac{1}{x^{2} + 1}$ is $\frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{2}$ .
      
      So, the result is: $7 \operatorname{atan}{\left(\frac{x}{2} \right)}$
      
      The result is: $\frac{9 \log{\left(x^{2} + 4 \right)}}{2} + 7 \operatorname{atan}{\left(\frac{x}{2} \right)}$
    So, the result is: $- \frac{9 \log{\left(x^{2} + 4 \right)}}{16} - \frac{7 \operatorname{atan}{\left(\frac{x}{2} \right)}}{8}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{17}{8 \left(x + 2\right)}\, dx = \frac{17 \int \frac{1}{x + 2}\, dx}{8}$
    1. Let $u = x + 2$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x + 2 \right)}$
    So, the result is: $\frac{17 \log{\left(x + 2 \right)}}{8}$
  The result is: $\frac{17 \log{\left(x + 2 \right)}}{8} - \frac{9 \log{\left(x^{2} + 4 \right)}}{16} - \frac{7 \operatorname{atan}{\left(\frac{x}{2} \right)}}{8}$
Method #2
1. Rewrite the integrand:
  
  $\frac{x^{2} - 4 x + 5}{\left(x + 2\right) \left(x^{2} + 4\right)} = \frac{x^{2} - 4 x + 5}{x^{3} + 2 x^{2} + 4 x + 8}$
2. Rewrite the integrand:
  
  $\frac{x^{2} - 4 x + 5}{x^{3} + 2 x^{2} + 4 x + 8} = - \frac{9 x + 14}{8 \left(x^{2} + 4\right)} + \frac{17}{8 \left(x + 2\right)}$
3. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{9 x + 14}{8 \left(x^{2} + 4\right)}\right)\, dx = - \frac{\int \frac{9 x + 14}{x^{2} + 4}\, dx}{8}$
    1. Rewrite the integrand:
      
      $\frac{9 x + 14}{x^{2} + 4} = \frac{9 x}{x^{2} + 4} + \frac{14}{x^{2} + 4}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{9 x}{x^{2} + 4}\, dx = \frac{9 \int \frac{2 x}{x^{2} + 4}\, dx}{2}$
      
      Let $u = x^{2} + 4$ .
      
      Then let $du = 2 x dx$ and substitute $\frac{9 du}{2}$ :
      
      $\int \frac{9}{2 u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x^{2} + 4 \right)}$
      
      So, the result is: $\frac{9 \log{\left(x^{2} + 4 \right)}}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{14}{x^{2} + 4}\, dx = 14 \int \frac{1}{x^{2} + 4}\, dx$
      
      The integral of $\frac{1}{x^{2} + 1}$ is $\frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{2}$ .
      
      So, the result is: $7 \operatorname{atan}{\left(\frac{x}{2} \right)}$
      
      The result is: $\frac{9 \log{\left(x^{2} + 4 \right)}}{2} + 7 \operatorname{atan}{\left(\frac{x}{2} \right)}$
    So, the result is: $- \frac{9 \log{\left(x^{2} + 4 \right)}}{16} - \frac{7 \operatorname{atan}{\left(\frac{x}{2} \right)}}{8}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{17}{8 \left(x + 2\right)}\, dx = \frac{17 \int \frac{1}{x + 2}\, dx}{8}$
    1. Let $u = x + 2$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x + 2 \right)}$
    So, the result is: $\frac{17 \log{\left(x + 2 \right)}}{8}$
  The result is: $\frac{17 \log{\left(x + 2 \right)}}{8} - \frac{9 \log{\left(x^{2} + 4 \right)}}{16} - \frac{7 \operatorname{atan}{\left(\frac{x}{2} \right)}}{8}$
Method #3
1. Rewrite the integrand:
  
  $\frac{x^{2} - 4 x + 5}{\left(x + 2\right) \left(x^{2} + 4\right)} = \frac{x^{2}}{x^{3} + 2 x^{2} + 4 x + 8} - \frac{4 x}{x^{3} + 2 x^{2} + 4 x + 8} + \frac{5}{x^{3} + 2 x^{2} + 4 x + 8}$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\frac{x^{2}}{x^{3} + 2 x^{2} + 4 x + 8} = \frac{x - 2}{2 \left(x^{2} + 4\right)} + \frac{1}{2 \left(x + 2\right)}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{x - 2}{2 \left(x^{2} + 4\right)}\, dx = \frac{\int \frac{x - 2}{x^{2} + 4}\, dx}{2}$
      
      Rewrite the integrand:
      
      $\frac{x - 2}{x^{2} + 4} = \frac{x}{x^{2} + 4} - \frac{2}{x^{2} + 4}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{x}{x^{2} + 4}\, dx = \frac{\int \frac{2 x}{x^{2} + 4}\, dx}{2}$
      
      Let $u = x^{2} + 4$ .
      
      Then let $du = 2 x dx$ 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:
      
      $\log{\left(x^{2} + 4 \right)}$
      
      So, the result is: $\frac{\log{\left(x^{2} + 4 \right)}}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{2}{x^{2} + 4}\right)\, dx = - 2 \int \frac{1}{x^{2} + 4}\, dx$
      
      The integral of $\frac{1}{x^{2} + 1}$ is $\frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{2}$ .
      
      So, the result is: $- \operatorname{atan}{\left(\frac{x}{2} \right)}$
      
      The result is: $\frac{\log{\left(x^{2} + 4 \right)}}{2} - \operatorname{atan}{\left(\frac{x}{2} \right)}$
      
      So, the result is: $\frac{\log{\left(x^{2} + 4 \right)}}{4} - \frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{2}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{2 \left(x + 2\right)}\, dx = \frac{\int \frac{1}{x + 2}\, dx}{2}$
      
      Let $u = x + 2$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x + 2 \right)}$
      
      So, the result is: $\frac{\log{\left(x + 2 \right)}}{2}$
    The result is: $\frac{\log{\left(x + 2 \right)}}{2} + \frac{\log{\left(x^{2} + 4 \right)}}{4} - \frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{4 x}{x^{3} + 2 x^{2} + 4 x + 8}\right)\, dx = - 4 \int \frac{x}{x^{3} + 2 x^{2} + 4 x + 8}\, dx$
    1. Rewrite the integrand:
      
      $\frac{x}{x^{3} + 2 x^{2} + 4 x + 8} = \frac{x + 2}{4 \left(x^{2} + 4\right)} - \frac{1}{4 \left(x + 2\right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{x + 2}{4 \left(x^{2} + 4\right)}\, dx = \frac{\int \frac{x + 2}{x^{2} + 4}\, dx}{4}$
      
      Rewrite the integrand:
      
      $\frac{x + 2}{x^{2} + 4} = \frac{x}{x^{2} + 4} + \frac{2}{x^{2} + 4}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{x}{x^{2} + 4}\, dx = \frac{\int \frac{2 x}{x^{2} + 4}\, dx}{2}$
      
      Let $u = x^{2} + 4$ .
      
      Then let $du = 2 x dx$ 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:
      
      $\log{\left(x^{2} + 4 \right)}$
      
      So, the result is: $\frac{\log{\left(x^{2} + 4 \right)}}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{2}{x^{2} + 4}\, dx = 2 \int \frac{1}{x^{2} + 4}\, dx$
      
      The integral of $\frac{1}{x^{2} + 1}$ is $\frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{2}$ .
      
      So, the result is: $\operatorname{atan}{\left(\frac{x}{2} \right)}$
      
      The result is: $\frac{\log{\left(x^{2} + 4 \right)}}{2} + \operatorname{atan}{\left(\frac{x}{2} \right)}$
      
      So, the result is: $\frac{\log{\left(x^{2} + 4 \right)}}{8} + \frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{4}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{4 \left(x + 2\right)}\right)\, dx = - \frac{\int \frac{1}{x + 2}\, dx}{4}$
      
      Let $u = x + 2$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x + 2 \right)}$
      
      So, the result is: $- \frac{\log{\left(x + 2 \right)}}{4}$
      
      The result is: $- \frac{\log{\left(x + 2 \right)}}{4} + \frac{\log{\left(x^{2} + 4 \right)}}{8} + \frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{4}$
    So, the result is: $\log{\left(x + 2 \right)} - \frac{\log{\left(x^{2} + 4 \right)}}{2} - \operatorname{atan}{\left(\frac{x}{2} \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{5}{x^{3} + 2 x^{2} + 4 x + 8}\, dx = 5 \int \frac{1}{x^{3} + 2 x^{2} + 4 x + 8}\, dx$
    1. Rewrite the integrand:
      
      $\frac{1}{x^{3} + 2 x^{2} + 4 x + 8} = - \frac{x - 2}{8 \left(x^{2} + 4\right)} + \frac{1}{8 \left(x + 2\right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{x - 2}{8 \left(x^{2} + 4\right)}\right)\, dx = - \frac{\int \frac{x - 2}{x^{2} + 4}\, dx}{8}$
      
      Rewrite the integrand:
      
      $\frac{x - 2}{x^{2} + 4} = \frac{x}{x^{2} + 4} - \frac{2}{x^{2} + 4}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{x}{x^{2} + 4}\, dx = \frac{\int \frac{2 x}{x^{2} + 4}\, dx}{2}$
      
      Let $u = x^{2} + 4$ .
      
      Then let $du = 2 x dx$ 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:
      
      $\log{\left(x^{2} + 4 \right)}$
      
      So, the result is: $\frac{\log{\left(x^{2} + 4 \right)}}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{2}{x^{2} + 4}\right)\, dx = - 2 \int \frac{1}{x^{2} + 4}\, dx$
      
      The integral of $\frac{1}{x^{2} + 1}$ is $\frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{2}$ .
      
      So, the result is: $- \operatorname{atan}{\left(\frac{x}{2} \right)}$
      
      The result is: $\frac{\log{\left(x^{2} + 4 \right)}}{2} - \operatorname{atan}{\left(\frac{x}{2} \right)}$
      
      So, the result is: $- \frac{\log{\left(x^{2} + 4 \right)}}{16} + \frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{8}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{8 \left(x + 2\right)}\, dx = \frac{\int \frac{1}{x + 2}\, dx}{8}$
      
      Let $u = x + 2$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(x + 2 \right)}$
      
      So, the result is: $\frac{\log{\left(x + 2 \right)}}{8}$
      
      The result is: $\frac{\log{\left(x + 2 \right)}}{8} - \frac{\log{\left(x^{2} + 4 \right)}}{16} + \frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{8}$
    So, the result is: $\frac{5 \log{\left(x + 2 \right)}}{8} - \frac{5 \log{\left(x^{2} + 4 \right)}}{16} + \frac{5 \operatorname{atan}{\left(\frac{x}{2} \right)}}{8}$
  The result is: $\frac{17 \log{\left(x + 2 \right)}}{8} - \frac{9 \log{\left(x^{2} + 4 \right)}}{16} - \frac{7 \operatorname{atan}{\left(\frac{x}{2} \right)}}{8}$
Add the constant of integration:

$\frac{17 \log{\left(x + 2 \right)}}{8} - \frac{9 \log{\left(x^{2} + 4 \right)}}{16} - \frac{7 \operatorname{atan}{\left(\frac{x}{2} \right)}}{8}+ \mathrm{constant}$

The answer is:

$\frac{17 \log{\left(x + 2 \right)}}{8} - \frac{9 \log{\left(x^{2} + 4 \right)}}{16} - \frac{7 \operatorname{atan}{\left(\frac{x}{2} \right)}}{8}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                   
 |                                                 /x\                
 |    2                           /     2\   7*atan|-|                
 |   x  - 4*x + 5            9*log\4 + x /         \2/   17*log(2 + x)
 | ---------------- dx = C - ------------- - --------- + -------------
 |         / 2    \                16            8             8      
 | (x + 2)*\x  + 4/                                                   
 |                                                                    
/

$$-{{9\,\log \left(x^2+4\right)}\over{16}}+{{17\,\log \left(x+2 \right)}\over{8}}-{{7\,\arctan \left({{x}\over{2}}\right)}\over{8}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  17*log(2)   9*log(5)   7*atan(1/2)   9*log(4)   17*log(3)
- --------- - -------- - ----------- + -------- + ---------
      8          16           8           16          8

$$-{{9\,\log 5}\over{16}}+{{9\,\log 4-34\,\log 2}\over{16}}+{{17\, \log 3}\over{8}}-{{7\,\arctan \left({{1}\over{2}}\right)}\over{8}}$$

  17*log(2)   9*log(5)   7*atan(1/2)   9*log(4)   17*log(3)
- --------- - -------- - ----------- + -------- + ---------
      8          16           8           16          8

$$- \frac{17 \log{\left(2 \right)}}{8} - \frac{9 \log{\left(5 \right)}}{16} - \frac{7 \operatorname{atan}{\left(\frac{1}{2} \right)}}{8} + \frac{9 \log{\left(4 \right)}}{16} + \frac{17 \log{\left(3 \right)}}{8}$$

Simplify

Numerical answer [src]

0.330403449239901

0.330403449239901

The graph

Integral of (x^2-4x+5)/((x+2)(x^2+4)) dx

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (x^2-4x+5)/((x+2)(x^2+4)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3