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Integral of d{x}:
Integral of e^(x^2)
Integral of e^(-x)
Integral of sin(x)^3
Integral of cos(x/2)
Identical expressions
(t^ two - one)/(t^(three)+t^ two)
(t squared minus 1) divide by (t to the power of (3) plus t squared )
(t to the power of two minus one) divide by (t to the power of (three) plus t to the power of two)
(t2-1)/(t(3)+t2)
t2-1/t3+t2
(t²-1)/(t^(3)+t²)
(t to the power of 2-1)/(t to the power of (3)+t to the power of 2)
t^2-1/t^3+t^2
(t^2-1) divide by (t^(3)+t^2)
Similar expressions
(t^2+1)/(t^(3)+t^2)
(t^2-1)/(t^(3)-t^2)

Integral of (t^2-1)/(t^(3)+t^2) dx

The solution

You have entered [src]

  1           
  /           
 |            
 |    2       
 |   t  - 1   
 |  ------- dt
 |   3    2   
 |  t  + t    
 |            
/             
0

$$\int\limits_{0}^{1} \frac{t^{2} - 1}{t^{3} + t^{2}}\, dt$$

Integral((t^2 - 1)/(t^3 + t^2), (t, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{t^{2} - 1}{t^{3} + t^{2}} = \frac{1}{t} - \frac{1}{t^{2}}$
2. Integrate term-by-term:
  1. The integral of $\frac{1}{t}$ is $\log{\left(t \right)}$ .
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{t^{2}}\right)\, dt = - \int \frac{1}{t^{2}}\, dt$
    1. The integral of $t^{n}$ is $\frac{t^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{t^{2}}\, dt = - \frac{1}{t}$
    So, the result is: $\frac{1}{t}$
  The result is: $\log{\left(t \right)} + \frac{1}{t}$
Method #2
1. Rewrite the integrand:
  
  $\frac{t^{2} - 1}{t^{3} + t^{2}} = \frac{t - 1}{t^{2}}$
2. Rewrite the integrand:
  
  $\frac{t - 1}{t^{2}} = \frac{1}{t} - \frac{1}{t^{2}}$
3. Integrate term-by-term:
  1. The integral of $\frac{1}{t}$ is $\log{\left(t \right)}$ .
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{t^{2}}\right)\, dt = - \int \frac{1}{t^{2}}\, dt$
    1. The integral of $t^{n}$ is $\frac{t^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{t^{2}}\, dt = - \frac{1}{t}$
    So, the result is: $\frac{1}{t}$
  The result is: $\log{\left(t \right)} + \frac{1}{t}$
Method #3
1. Rewrite the integrand:
  
  $\frac{t^{2} - 1}{t^{3} + t^{2}} = \frac{t^{2}}{t^{3} + t^{2}} - \frac{1}{t^{3} + t^{2}}$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\frac{t^{2}}{t^{3} + t^{2}} = \frac{1}{t + 1}$
  2. Let $u = t + 1$ .
    
    Then let $du = dt$ and substitute $du$ :
    
    $\int \frac{1}{u}\, du$
    1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
    Now substitute $u$ back in:
    
    $\log{\left(t + 1 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{t^{3} + t^{2}}\right)\, dt = - \int \frac{1}{t^{3} + t^{2}}\, dt$
    1. Rewrite the integrand:
      
      $\frac{1}{t^{3} + t^{2}} = \frac{1}{t + 1} - \frac{1}{t} + \frac{1}{t^{2}}$
    2. Integrate term-by-term:
      
      Let $u = t + 1$ .
      
      Then let $du = dt$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(t + 1 \right)}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{t}\right)\, dt = - \int \frac{1}{t}\, dt$
      
      The integral of $\frac{1}{t}$ is $\log{\left(t \right)}$ .
      
      So, the result is: $- \log{\left(t \right)}$
      
      The integral of $t^{n}$ is $\frac{t^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{t^{2}}\, dt = - \frac{1}{t}$
      
      The result is: $- \log{\left(t \right)} + \log{\left(t + 1 \right)} - \frac{1}{t}$
    So, the result is: $\log{\left(t \right)} - \log{\left(t + 1 \right)} + \frac{1}{t}$
  The result is: $\log{\left(t \right)} + \frac{1}{t}$
Add the constant of integration:

$\log{\left(t \right)} + \frac{1}{t}+ \mathrm{constant}$

The answer is:

$\log{\left(t \right)} + \frac{1}{t}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                           
 |                            
 |   2                        
 |  t  - 1          1         
 | ------- dt = C + - + log(t)
 |  3    2          t         
 | t  + t                     
 |                            
/

$$\int \frac{t^{2} - 1}{t^{3} + t^{2}}\, dt = C + \log{\left(t \right)} + \frac{1}{t}$$

Simplify

The answer [src]

-oo

$$-\infty$$

-oo

$$-\infty$$

-oo

Numerical answer [src]

-1.3793236779486e+19

-1.3793236779486e+19

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 (t^2-1)/(t^(3)+t^2) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3