What is the integral of 1/(y-x) ?

The integral of 1/(y-x) is -ln|y-x|+C

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integral of 1/(y-x)

\int \frac{1}{y - x} d x

- ln ∣ y - x ∣ + C

Solution steps

\int \frac{1}{y - x} d x

Apply u-substitution

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

= \int - \frac{1}{u} d u

Take the constant out:

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

= - \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = y - x

= - ln ∣ y - x ∣

Add a constant to the solution

= - ln ∣ y - x ∣ + C

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x \to - 4 lim (x^{2} - 16)

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