Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of x/(x^2+3)
Integral of x*e^(5*x)*dx
Integral of (xcosx)/(2+3x^3)
Integral of x^4/(x^4-2*x^2+1)
Identical expressions
(7x+ five)*lnx
(7x plus 5) multiply by lnx
(7x plus five) multiply by lnx
(7x+5)lnx
7x+5lnx
(7x+5)*lnxdx
Similar expressions
(7x-5)*lnx

Integral of (7x+5)*lnx dx

The solution

You have entered [src]

  1                    
  /                    
 |                     
 |  (7*x + 5)*log(x) dx
 |                     
/                      
0

$$\int\limits_{0}^{1} \left(7 x + 5\right) \log{\left(x \right)}\, dx$$

Integral((7*x + 5)*log(x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = \log{\left(x \right)}$ .
  
  Then let $du = \frac{dx}{x}$ and substitute $du$ :
  
  $\int \left(7 u e^{2 u} + 5 u e^{u}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 7 u e^{2 u}\, du = 7 \int u e^{2 u}\, du$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      So, the result is: $\frac{e^{2 u}}{4}$
      
      So, the result is: $\frac{7 u e^{2 u}}{2} - \frac{7 e^{2 u}}{4}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 5 u e^{u}\, du = 5 \int u e^{u}\, du$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      Now evaluate the sub-integral.
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $5 u e^{u} - 5 e^{u}$
    The result is: $\frac{7 u e^{2 u}}{2} + 5 u e^{u} - \frac{7 e^{2 u}}{4} - 5 e^{u}$
  Now substitute $u$ back in:
  
  $\frac{7 x^{2} \log{\left(x \right)}}{2} - \frac{7 x^{2}}{4} + 5 x \log{\left(x \right)} - 5 x$
Method #2
1. Rewrite the integrand:
  
  $\left(7 x + 5\right) \log{\left(x \right)} = 7 x \log{\left(x \right)} + 5 \log{\left(x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 7 x \log{\left(x \right)}\, dx = 7 \int x \log{\left(x \right)}\, dx$
    1. Let $u = \log{\left(x \right)}$ .
      
      Then let $du = \frac{dx}{x}$ and substitute $du$ :
      
      $\int u e^{2 u}\, du$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      So, the result is: $\frac{e^{2 u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}$
    So, the result is: $\frac{7 x^{2} \log{\left(x \right)}}{2} - \frac{7 x^{2}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 5 \log{\left(x \right)}\, dx = 5 \int \log{\left(x \right)}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = \log{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$ .
      
      Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$ .
      
      To find $v{\left(x \right)}$ :
      
      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 is the constant times the variable of integration:
      
      $\int 1\, dx = x$
    So, the result is: $5 x \log{\left(x \right)} - 5 x$
  The result is: $\frac{7 x^{2} \log{\left(x \right)}}{2} - \frac{7 x^{2}}{4} + 5 x \log{\left(x \right)} - 5 x$
Method #3
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = \log{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 7 x + 5$ .
  
  Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$ .
  
  To find $v{\left(x \right)}$ :
  1. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 7 x\, dx = 7 \int x\, dx$
      
      The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
      
      So, the result is: $\frac{7 x^{2}}{2}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int 5\, dx = 5 x$
    The result is: $\frac{7 x^{2}}{2} + 5 x$
  Now evaluate the sub-integral.
2. Rewrite the integrand:
  
  $\frac{\frac{7 x^{2}}{2} + 5 x}{x} = \frac{7 x}{2} + 5$
3. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{7 x}{2}\, dx = \frac{7 \int x\, dx}{2}$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
    So, the result is: $\frac{7 x^{2}}{4}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 5\, dx = 5 x$
  The result is: $\frac{7 x^{2}}{4} + 5 x$
Method #4
1. Rewrite the integrand:
  
  $\left(7 x + 5\right) \log{\left(x \right)} = 7 x \log{\left(x \right)} + 5 \log{\left(x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 7 x \log{\left(x \right)}\, dx = 7 \int x \log{\left(x \right)}\, dx$
    1. Let $u = \log{\left(x \right)}$ .
      
      Then let $du = \frac{dx}{x}$ and substitute $du$ :
      
      $\int u e^{2 u}\, du$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      So, the result is: $\frac{e^{2 u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}$
    So, the result is: $\frac{7 x^{2} \log{\left(x \right)}}{2} - \frac{7 x^{2}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 5 \log{\left(x \right)}\, dx = 5 \int \log{\left(x \right)}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = \log{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$ .
      
      Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$ .
      
      To find $v{\left(x \right)}$ :
      
      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 is the constant times the variable of integration:
      
      $\int 1\, dx = x$
    So, the result is: $5 x \log{\left(x \right)} - 5 x$
  The result is: $\frac{7 x^{2} \log{\left(x \right)}}{2} - \frac{7 x^{2}}{4} + 5 x \log{\left(x \right)} - 5 x$
Now simplify:

$\frac{x \left(14 x \log{\left(x \right)} - 7 x + 20 \log{\left(x \right)} - 20\right)}{4}$
Add the constant of integration:

$\frac{x \left(14 x \log{\left(x \right)} - 7 x + 20 \log{\left(x \right)} - 20\right)}{4}+ \mathrm{constant}$

The answer is:

$\frac{x \left(14 x \log{\left(x \right)} - 7 x + 20 \log{\left(x \right)} - 20\right)}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                   2                   2       
 |                                 7*x                 7*x *log(x)
 | (7*x + 5)*log(x) dx = C - 5*x - ---- + 5*x*log(x) + -----------
 |                                  4                       2     
/

$$\int \left(7 x + 5\right) \log{\left(x \right)}\, dx = C + \frac{7 x^{2} \log{\left(x \right)}}{2} - \frac{7 x^{2}}{4} + 5 x \log{\left(x \right)} - 5 x$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

-27/4

$$- \frac{27}{4}$$

-27/4

$$- \frac{27}{4}$$

-27/4

Numerical answer [src]

-6.75

-6.75

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (7x+5)*lnx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4