What is the integral of 1/((60-x)(150-x)) ?

The integral of 1/((60-x)(150-x)) is -1/90 ln|60-x|+1/90 ln|-x+150|+C

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integral of 1/((60-x)(150-x))

\int \frac{1}{( 60 - x ) ( 150 - x )} d x

- \frac{1}{90} ln ∣ 60 - x ∣ + \frac{1}{90} ln ∣ - x + 150 ∣ + C

Solution steps

\int \frac{1}{( 60 - x ) ( 150 - x )} d x

Apply u-substitution

: \int - \frac{1}{u ( u + 90 )} d u

= \int - \frac{1}{u ( u + 90 )} 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 + 90 )} d u

Take the partial fraction of

\frac{1}{u ( u + 90 )} : \frac{1}{90 u} - \frac{1}{90 ( u + 90 )}

= - \int \frac{1}{90 u} - \frac{1}{90 ( u + 90 )} 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}{90 u} d u - \int \frac{1}{90 ( u + 90 )} d u)

\int \frac{1}{90 u} d u = \frac{1}{90} ln ∣ u ∣

\int \frac{1}{90 ( u + 90 )} d u = \frac{1}{90} ln ∣ u + 90 ∣

= - (\frac{1}{90} ln ∣ u ∣ - \frac{1}{90} ln ∣ u + 90 ∣)

Substitute back

u = 60 - x

= - (\frac{1}{90} ln ∣ 60 - x ∣ - \frac{1}{90} ln ∣ 60 - x + 90 ∣)

Simplify

- (\frac{1}{90} ln ∣ 60 - x ∣ - \frac{1}{90} ln ∣ 60 - x + 90 ∣) : - \frac{1}{90} ln ∣ 60 - x ∣ + \frac{1}{90} ln ∣ - x + 150 ∣

= - \frac{1}{90} ln ∣ 60 - x ∣ + \frac{1}{90} ln ∣ - x + 150 ∣

Add a constant to the solution

= - \frac{1}{90} ln ∣ 60 - x ∣ + \frac{1}{90} ln ∣ - x + 150 ∣ + C

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What is the integral of 1/((60-x)(150-x)) ?
The integral of 1/((60-x)(150-x)) is -1/90 ln|60-x|+1/90 ln|-x+150|+C