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

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

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integral of x/(1+ax)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u - 1}{a ^{2} u} d u

= \int \frac{u - 1}{a ^{2} u} d u

Take the constant out:

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

= \frac{1}{a ^{2}} \cdot \int \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= \frac{1}{a ^{2}} \cdot \int 1 - \frac{1}{u} d u

Apply the Sum Rule:

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

= \frac{1}{a ^{2}} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{a ^{2}} (u - ln ∣ u ∣)

Substitute back

u = 1 + a x

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

Add a constant to the solution

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

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