Find the indefinite integral of 1/(a - bx) with respect to x, where a and b are constants and b is not zero.

Step-by-step solution

1. 1

   Choose the substitution. The denominator is linear, so let the whole denominator be the new variable:

   $u=a-bx$

   This is the standard move for any $\int\frac{dx}{\text{linear}}$ — do not reach for partial fractions or a $\ln$ formula from memory.

2. 2

   Differentiate the substitution and solve for dx.

   $du=-b\,dx\qquad\Longrightarrow\qquad dx=-\frac{1}{b}\,du$

   The minus sign comes from the $-bx$; losing it is the most common error in this integral.

3. 3

   Rewrite the integral in u.

   $\int\frac{1}{a-bx}\,dx=\int\frac{1}{u}\left(-\frac{1}{b}\right)du=-\frac{1}{b}\int\frac{du}{u}$

   The constant $-\tfrac1b$ pulls straight out of the integral.

4. 4

   Integrate and substitute back. Since $\int\frac{du}{u}=\ln|u|+C$:

   $-\frac{1}{b}\ln|u|+C=-\frac{1}{b}\ln|a-bx|+C$

   The absolute value is essential: $a-bx$ changes sign at $x=\tfrac{a}{b}$, where the integrand has a vertical asymptote, and the antiderivative is valid on each side of it separately.

5. 5

   Note the general pattern. For any linear inner function,

   $\int\frac{dx}{px+q}=\frac{1}{p}\ln|px+q|+C$

   Here $p=-b$, which reproduces the $-\tfrac1b$ directly. The coefficient out front is always the reciprocal of the inner slope.

6. 6

   Verify by differentiating. By the chain rule, $\frac{d}{dx}\left[-\frac1b\ln|a-bx|\right]=-\frac1b\cdot\frac{-b}{a-bx}=\frac{1}{a-bx}$ ✓. Numerically with $a=3.7$, $b=1.9$, symmetric difference quotients at $x=0.4,\,-2.0,\,3.0$ match the integrand to five digits ✓, including a point on each side of the asymptote at $x=1.947$.