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The double integral of:
1/(x+y+1)
1/(x+y+1)
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1/x+y+1
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Similar expressions
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Integral of 1/(x+y+1) dy

The solution

You have entered [src]

  2             
  /             
 |              
 |      1       
 |  --------- dy
 |  x + y + 1   
 |              
/               
1

$$\int\limits_{1}^{2} \frac{1}{\left(x + y\right) + 1}\, dy$$

Integral(1/(x + y + 1), (y, 1, 2))

Detail solution

Let $u = \left(x + y\right) + 1$ .

Then let $du = dy$ 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:

$\log{\left(\left(x + y\right) + 1 \right)}$
Now simplify:

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

$\log{\left(x + y + 1 \right)}+ \mathrm{constant}$

The answer is:

$\log{\left(x + y + 1 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

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

Simplify

The answer [src]

-log(2 + x) + log(3 + x)

$$- \log{\left(x + 2 \right)} + \log{\left(x + 3 \right)}$$

-log(2 + x) + log(3 + x)

$$- \log{\left(x + 2 \right)} + \log{\left(x + 3 \right)}$$

-log(2 + x) + log(3 + x)

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

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Identical expressions

Similar expressions

Integral of 1/(x+y+1) dy

Limits of integration:

The graph:

Piecewise:

The solution