integrate 1/(x+1) — step-by-step solution

Let's integrate:

\int \frac{1}{x + 1} d x

Let

u = x + 1

. Then

du = dx

, so

dx = du

\int \frac{1}{x + 1} d x = \int \frac{1}{u} d u

We use

\int \frac{1}{u} d u = \ln | u |

. The absolute value covers both signs of the denominator, on any interval where

x + 1 \neq 0

Substitute

u

x + 1

back into the antiderivative

ⓘ The derivative of

\ln | x + 1 |

\frac{1}{x + 1}

, so

\ln | x + 1 |

is an antiderivative of

\frac{1}{x + 1}

\int \frac{1}{x + 1} d x

\ln | x + 1 |

\ln | x + 1 |

▶Indefinite integrals are defined up to an additive constant C, so this yields:

\ln | x + 1 | + C