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

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

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integral of 1/(x+ax)

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

\frac{1}{1 + a} ln ∣ x + a x ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ( 1 + a )} d u

= \int \frac{1}{u ( 1 + a )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{1 + a} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{1 + a} ln ∣ u ∣

Substitute back

u = x + a x

= \frac{1}{1 + a} ln ∣ x + a x ∣

Add a constant to the solution

= \frac{1}{1 + a} ln ∣ x + a x ∣ + C

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