What is the integral of (ax)/(x^2-bx) ?

The integral of (ax)/(x^2-bx) is aln|x-b|+C

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integral of (ax)/(x^2-bx)

\int \frac{a x}{x ^{2} - b x} d x

a ln ∣ x - b ∣ + C

Solution steps

\int \frac{a x}{x ^{2} - b x} d x

Take the constant out:

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

= a \cdot \int \frac{x}{x ^{2} - b x} d x

Simplify

\frac{x}{x ^{2} - b x} : \frac{1}{x - b}

= a \cdot \int \frac{1}{x - b} d x

Apply u-substitution

: \int \frac{1}{u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= a ln ∣ u ∣

Substitute back

u = x - b

= a ln ∣ x - b ∣

Add a constant to the solution

= a ln ∣ x - b ∣ + C

\frac{d y}{d t} = \frac{3 t ^{2} + t e ^{- t} + 1}{2 y - 4}

\int \frac{a}{x ^{2}} d x

\int \frac{1}{( x + 3 )} d x

\frac{d}{d x} (\frac{( 3 - 2 x - x ^{2} )}{( x ^{2} - 1 )})

\frac{d y}{d t} = y \cdot ln (t), y (e) = 200

What is the integral of (ax)/(x^2-bx) ?
The integral of (ax)/(x^2-bx) is aln|x-b|+C