What is the integral of 1/((15-x)(60-x)) ?

The integral of 1/((15-x)(60-x)) is -1/45 ln|15-x|+1/45 ln|-x+60|+C

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integral of 1/((15-x)(60-x))

\int \frac{1}{( 15 - x ) ( 60 - x )} d x

- \frac{1}{45} ln ∣ 15 - x ∣ + \frac{1}{45} ln ∣ - x + 60 ∣ + C

Solution steps

\int \frac{1}{( 15 - x ) ( 60 - x )} d x

Apply u-substitution

: \int - \frac{1}{u ( u + 45 )} d u

= \int - \frac{1}{u ( u + 45 )} d u

Take the constant out:

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

= - \int \frac{1}{u ( u + 45 )} d u

Take the partial fraction of

\frac{1}{u ( u + 45 )} : \frac{1}{45 u} - \frac{1}{45 ( u + 45 )}

= - \int \frac{1}{45 u} - \frac{1}{45 ( u + 45 )} 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

= - (\int \frac{1}{45 u} d u - \int \frac{1}{45 ( u + 45 )} d u)

\int \frac{1}{45 u} d u = \frac{1}{45} ln ∣ u ∣

\int \frac{1}{45 ( u + 45 )} d u = \frac{1}{45} ln ∣ u + 45 ∣

= - (\frac{1}{45} ln ∣ u ∣ - \frac{1}{45} ln ∣ u + 45 ∣)

Substitute back

u = 15 - x

= - (\frac{1}{45} ln ∣ 15 - x ∣ - \frac{1}{45} ln ∣ 15 - x + 45 ∣)

Simplify

- (\frac{1}{45} ln ∣ 15 - x ∣ - \frac{1}{45} ln ∣ 15 - x + 45 ∣) : - \frac{1}{45} ln ∣ 15 - x ∣ + \frac{1}{45} ln ∣ - x + 60 ∣

= - \frac{1}{45} ln ∣ 15 - x ∣ + \frac{1}{45} ln ∣ - x + 60 ∣

Add a constant to the solution

= - \frac{1}{45} ln ∣ 15 - x ∣ + \frac{1}{45} ln ∣ - x + 60 ∣ + C

x \to - 1 lim (\frac{k x + 3}{4})

y = x^{3} - 3 x^{2} + 4

\frac{d}{d x} (2 (6 - x^{2})^{5})

\frac{\partial}{\partial x} (sin (x) \cdot cos (y))

x \to 1 lim (x (\frac{10}{x - 1}))

What is the integral of 1/((15-x)(60-x)) ?
The integral of 1/((15-x)(60-x)) is -1/45 ln|15-x|+1/45 ln|-x+60|+C