# Integral of $\frac{1}{x + 1}$

The calculator will find the integral/antiderivative of $\frac{1}{x + 1}$, with steps shown.

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Find $\int \frac{1}{x + 1}\, dx$.

### Solution

Let $u=x + 1$.

Then $du=\left(x + 1\right)^{\prime }dx = 1 dx$ (steps can be seen here), and we have that $dx = du$.

The integral becomes

$$\color{red}{\int{\frac{1}{x + 1} d x}} = \color{red}{\int{\frac{1}{u} d u}}$$

The integral of $\frac{1}{u}$ is $\int{\frac{1}{u} d u} = \ln{\left(u \right)}$

$$\color{red}{\int{\frac{1}{u} d u}} = \color{red}{\ln{\left(u \right)}}$$

Recall that $u=x + 1$:

$$\ln{\left(\color{red}{u} \right)} = \ln{\left(\color{red}{\left(x + 1\right)} \right)}$$

Therefore,

$$\int{\frac{1}{x + 1} d x} = \ln{\left(\left|{x + 1}\right| \right)}$$

$$\int{\frac{1}{x + 1} d x} = \ln{\left(\left|{x + 1}\right| \right)}+C$$
Answer: $\int{\frac{1}{x + 1} d x}=\ln{\left(\left|{x + 1}\right| \right)}+C$