What is the integral of (1/120)/(-P+120) ?

The integral of (1/120)/(-P+120) is -1/120 ln|-P+120|+C

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integral of (1/120)/(-P+120)

\int \frac{\frac{1}{120}}{- P + 120} d P

- \frac{1}{120} ln ∣ - P + 120 ∣ + C

Solution steps

\int \frac{\frac{1}{120}}{- P + 120} d P

Take the constant out:

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

= \frac{1}{120} \cdot \int \frac{1}{- P + 120} d P

Apply u-substitution

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

= \frac{1}{120} \cdot \int - \frac{1}{u} d u

Take the constant out:

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

= \frac{1}{120} (- \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{120} (- ln ∣ u ∣)

Substitute back

u = - P + 120

= \frac{1}{120} (- ln ∣ - P + 120 ∣)

Simplify

= - \frac{1}{120} ln ∣ - P + 120 ∣

Add a constant to the solution

= - \frac{1}{120} ln ∣ - P + 120 ∣ + C

tan (sin (x))

\frac{d y}{d x} = \frac{( e ^{x} - e ^{- x} )}{( 3 + 4 y )}

y = x - 1, y^{2} = 2 x + 6

\int 4 e^{x} 10 + e^{x} d x

\frac{1 - 5 t}{6 + t}

What is the integral of (1/120)/(-P+120) ?
The integral of (1/120)/(-P+120) is -1/120 ln|-P+120|+C