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Identical expressions
ln^ two (x+ three)
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ln2(x+3)
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Similar expressions
ln^2(x-3)

Integral of ln^2(x+3) dx

The solution

You have entered [src]

 -1 + E              
    /                
   |                 
   |      2          
   |   log (x + 3) dx
   |                 
  /                  
  0

$$\int\limits_{0}^{-1 + e} \log{\left(x + 3 \right)}^{2}\, dx$$

Integral(log(x + 3)^2, (x, 0, -1 + E))

Detail solution

Let $u = \log{\left(x + 3 \right)}$ .

Then let $du = \frac{dx}{x + 3}$ and substitute $du$ :

$\int u^{2} e^{u}\, du$
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(u \right)} = u^{2}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$ .
  
  Then $\operatorname{du}{\left(u \right)} = 2 u$ .
  
  To find $v{\left(u \right)}$ :
  1. The integral of the exponential function is itself.
    
    $\int e^{u}\, du = e^{u}$
  Now evaluate the sub-integral.
2. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(u \right)} = 2 u$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$ .
  
  Then $\operatorname{du}{\left(u \right)} = 2$ .
  
  To find $v{\left(u \right)}$ :
  1. The integral of the exponential function is itself.
    
    $\int e^{u}\, du = e^{u}$
  Now evaluate the sub-integral.
3. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 2 e^{u}\, du = 2 \int e^{u}\, du$
  1. The integral of the exponential function is itself.
    
    $\int e^{u}\, du = e^{u}$
  So, the result is: $2 e^{u}$
Now substitute $u$ back in:

$2 x + \left(x + 3\right) \log{\left(x + 3 \right)}^{2} - 2 \left(x + 3\right) \log{\left(x + 3 \right)} + 6$
Now simplify:

$2 x + \left(x + 3\right) \log{\left(x + 3 \right)}^{2} - 2 \left(x + 3\right) \log{\left(x + 3 \right)} + 6$
Add the constant of integration:

$2 x + \left(x + 3\right) \log{\left(x + 3 \right)}^{2} - 2 \left(x + 3\right) \log{\left(x + 3 \right)} + 6+ \mathrm{constant}$

The answer is:

$2 x + \left(x + 3\right) \log{\left(x + 3 \right)}^{2} - 2 \left(x + 3\right) \log{\left(x + 3 \right)} + 6+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                         
 |                                                                          
 |    2                              2                                      
 | log (x + 3) dx = 6 + C + 2*x + log (x + 3)*(x + 3) - 2*(x + 3)*log(x + 3)
 |                                                                          
/

$$\int \log{\left(x + 3 \right)}^{2}\, dx = C + 2 x + \left(x + 3\right) \log{\left(x + 3 \right)}^{2} - 2 \left(x + 3\right) \log{\left(x + 3 \right)} + 6$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                         2                          2                                       
-2 - 6*log(2 + E) - 3*log (3) + 2*E + 6*log(3) + log (2 + E)*(2 + E) - 2*(-1 + E)*log(2 + E)

$$- 6 \log{\left(2 + e \right)} - 2 \left(-1 + e\right) \log{\left(2 + e \right)} - 3 \log{\left(3 \right)}^{2} - 2 + 2 e + 6 \log{\left(3 \right)} + \left(2 + e\right) \log{\left(2 + e \right)}^{2}$$

                         2                          2                                       
-2 - 6*log(2 + E) - 3*log (3) + 2*E + 6*log(3) + log (2 + E)*(2 + E) - 2*(-1 + E)*log(2 + E)

-2 - 6*log(2 + E) - 3*log(3)^2 + 2*E + 6*log(3) + log(2 + E)^2*(2 + E) - 2*(-1 + E)*log(2 + E)

Numerical answer [src]

3.12389700337105

3.12389700337105

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

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Integral of ln^2(x+3) dx

Limits of integration:

The graph:

Piecewise:

The solution