What is the integral of (a-x)/(1+ax) ?

The integral of (a-x)/(1+ax) is ln|1+ax|-1/(a^2)(1+ax-ln|1+ax|)+C

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integral of (a-x)/(1+ax)

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

ln ∣ 1 + a x ∣ - \frac{1}{a ^{2}} (1 + a x - ln ∣ 1 + a x ∣) + C

Solution steps

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

Expand

\frac{a - x}{1 + a x} : \frac{a}{1 + a x} - \frac{x}{1 + a x}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{a}{1 + a x} d x - \int \frac{x}{1 + a x} d x

\int \frac{a}{1 + a x} d x = ln ∣ 1 + a x ∣

\int \frac{x}{1 + a x} d x = \frac{1}{a ^{2}} (1 + a x - ln ∣ 1 + a x ∣)

= ln ∣ 1 + a x ∣ - \frac{1}{a ^{2}} (1 + a x - ln ∣ 1 + a x ∣)

Add a constant to the solution

= ln ∣ 1 + a x ∣ - \frac{1}{a ^{2}} (1 + a x - ln ∣ 1 + a x ∣) + C

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What is the integral of (a-x)/(1+ax) ?
The integral of (a-x)/(1+ax) is ln|1+ax|-1/(a^2)(1+ax-ln|1+ax|)+C