What is the integral of 1/(100-y) ?

The integral of 1/(100-y) is -ln|100-y|+C

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integral of 1/(100-y)

\int \frac{1}{100 - y} d y

- ln ∣ 100 - y ∣ + C

Solution steps

\int \frac{1}{100 - y} d y

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 = 100 - y

= - ln ∣ 100 - y ∣

Add a constant to the solution

= - ln ∣ 100 - y ∣ + C

\frac{\partial}{\partial x} (e^{x - 2} - \frac{y}{2})

\frac{d}{d x} (\frac{sin ( x )}{3 x})

\int 1.5 x^{2} d x

\frac{d y}{d x} + y^{5} x + 8 y = 0

\frac{π x}{16}