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The double integral of:
1/(x+y)
1/(x+y)
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one /(x+y)
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1/(x+y)dx
Similar expressions
1/(x-y)
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Integral of 1/(x+y) dx

The solution

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  1         
  /         
 |          
 |    1     
 |  ----- dx
 |  x + y   
 |          
/           
0

$$\int\limits_{0}^{1} \frac{1}{x + y}\, dx$$

Integral(1/(x + y), (x, 0, 1))

Detail solution

Let $u = x + y$ .

Then let $du = dx$ 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:

$\log{\left(x + y \right)}$
Add the constant of integration:

$\log{\left(x + y \right)}+ \mathrm{constant}$

The answer is:

$\log{\left(x + y \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                         
 |                          
 |   1                      
 | ----- dx = C + log(x + y)
 | x + y                    
 |                          
/

$$\int \frac{1}{x + y}\, dx = C + \log{\left(x + y \right)}$$

Simplify

The answer [src]

-log(y) + log(1 + y)

$$- \log{\left(y \right)} + \log{\left(y + 1 \right)}$$

-log(y) + log(1 + y)

$$- \log{\left(y \right)} + \log{\left(y + 1 \right)}$$

-log(y) + log(1 + y)

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

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Identical expressions

Similar expressions

Integral of 1/(x+y) dx

Limits of integration:

The graph:

Piecewise:

The solution