Integral of dx/(x+a) dx

The solution

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

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

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

Detail solution

Let $u = a + x$ .

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:

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The answer [src]

-log(a) + log(1 + a)

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

-log(a) + log(1 + a)

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

-log(a) + log(1 + a)

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

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Integral of dx/(x+a) dx

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The solution