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Integral of dx/(1+x^2)
Identical expressions
(e^(x+y)-y)dx+(xe^(x+y)+ one)
(e to the power of (x plus y) minus y)dx plus (xe to the power of (x plus y) plus 1)
(e to the power of (x plus y) minus y)dx plus (xe to the power of (x plus y) plus one)
(e(x+y)-y)dx+(xe(x+y)+1)
ex+y-ydx+xex+y+1
e^x+y-ydx+xe^x+y+1
Similar expressions
(e^(x-y)-y)dx+(xe^(x+y)+1)
(e^(x+y)-y)dx+(xe^(x-y)+1)
(e^(x+y)-y)dx-(xe^(x+y)+1)
(e^(x+y)+y)dx+(xe^(x+y)+1)
(e^(x+y)-y)dx+(xe^(x+y)-1)

Solving integrals/ (e^(x+y)-y)dx+(xe^(x+y)+1)

Integral of (e^(x+y)-y)dx+(xe^(x+y)+1) dy

The solution

You have entered [src]

  1                               
  /                               
 |                                
 |  / x + y          x + y    \   
 |  \E      - y + x*E      + 1/ dx
 |                                
/                                 
0

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

Integral(E^(x + y) - y + x*E^(x + y) + 1, (x, 0, 1))

Detail solution

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

$x \left(- y + e^{x + y} + 1\right)$
Add the constant of integration:

$x \left(- y + e^{x + y} + 1\right)+ \mathrm{constant}$

The answer is:

$x \left(- y + e^{x + y} + 1\right)+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                        
 |                                                                         
 | / x + y          x + y    \              /   x      x\  y          x + y
 | \E      - y + x*E      + 1/ dx = C + x + \- e  + x*e /*e  - x*y + e     
 |                                                                         
/

$$\int \left(\left(e^{x + y} - y\right) + \left(e^{x + y} x + 1\right)\right)\, dx = C - x y + x + \left(x e^{x} - e^{x}\right) e^{y} + e^{x + y}$$

Simplify

The answer [src]

         1 + y
1 - y + e

$$- y + e^{y + 1} + 1$$

         1 + y
1 - y + e

$$- y + e^{y + 1} + 1$$

1 - y + exp(1 + y)

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

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Identical expressions

Similar expressions

Integral of (e^(x+y)-y)dx+(xe^(x+y)+1) dy

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2