Mister Exam

Other calculators

Solving integrals/ 1/(x+y+1)

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

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
  2             
  /             
 |              
 |      1       
 |  --------- dy
 |  x + y + 1   
 |              
/               
1
∫121(x+y)+1 dy\int\limits_{1}^{2} \frac{1}{\left(x + y\right) + 1}\, dy 
Integral(1/(x + y + 1), (y, 1, 2))
Detail solution

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

    Then let du=dydu = dy and substitute dudu:

    ∫1u du\int \frac{1}{u}\, du

    1. The integral of 1u\frac{1}{u} is log⁡(u)\log{\left(u \right)}.

    Now substitute uu back in:

    log⁡((x+y)+1)\log{\left(\left(x + y\right) + 1 \right)}

  2. Now simplify:

    log⁡(x+y+1)\log{\left(x + y + 1 \right)}

  3. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                                 
 |                                  
 |     1                            
 | --------- dy = C + log(x + y + 1)
 | x + y + 1                        
 |                                  
/
∫1(x+y)+1 dy=C+log⁡((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⁡(x+2)+log⁡(x+3)- \log{\left(x + 2 \right)} + \log{\left(x + 3 \right)} 
=
= 
-log(2 + x) + log(3 + x)
−log⁡(x+2)+log⁡(x+3)- \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.

    © Mister Exam – Calculator