Integral de $$$\frac{1}{x + 1}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu aportación
Encuentra $$$\int \frac{1}{x + 1}\, dx$$$.
Solución
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 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$
$${\color{red}{\int{\frac{1}{u} d u}}} = {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$
Recall that $$$u=x + 1$$$:
$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\left(x + 1\right)}}}\right| \right)}$$
Therefore,
$$\int{\frac{1}{x + 1} d x} = \ln{\left(\left|{x + 1}\right| \right)}$$
Add the constant of integration:
$$\int{\frac{1}{x + 1} d x} = \ln{\left(\left|{x + 1}\right| \right)}+C$$
Answer: $$$\int{\frac{1}{x + 1} d x}=\ln{\left(\left|{x + 1}\right| \right)}+C$$$