Integral of 1÷(x+1) dx

The solution

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

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

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

Detail solution

Let $u = x + 1$ .

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:

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

log(2)

$$\log{\left(2 \right)}$$

log(2)

$$\log{\left(2 \right)}$$

log(2)

Numerical answer [src]

0.693147180559945

0.693147180559945

The graph

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

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Identical expressions

Similar expressions

Integral of 1÷(x+1) dx

Limits of integration:

The graph:

Piecewise:

The solution