integrate 1/(7x+2) — step-by-step solution

Let's compute the integral:

\int(\frac{1}{7 x + 2})dx

Let

u = 7 x + 2

. Then

du = 7 dx

, so

dx = \frac{du}{7}

\int(\frac{1}{7 x + 2})dx = \frac{1}{7} \cdot \int(\frac{1}{u})du

We use

\int(\frac{1}{u})du = \ln (abs (u))

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

7 x + 2!= 0

Substitute

u

7 x + 2

back into the antiderivative

\frac{1}{7} \cdot \ln (abs (7 x + 2))

\frac{\ln (abs (7 x + 2))}{7}

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

\frac{\ln (abs (7 x + 2))}{7} + C