What is the integral of 1/(P(a-bP)) ?

The integral of 1/(P(a-bP)) is 1/a (-ln((|a-bP|}{|a|})+ln|\frac{a-bP)/a-1|)+C

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integral of 1/(P(a-bP))

\int \frac{1}{P ( a - b P )} d P

\frac{1}{a} (- ln (\frac{∣ a - b P ∣}{∣ a ∣}) + ln \frac{a - b P}{a} - 1) + C

Solution steps

\int \frac{1}{P ( a - b P )} d P

Apply u-substitution

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

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

u - a = a (\frac{u}{a} - 1)

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{v ( v - 1 )} d v

= \frac{1}{a} \cdot \int \frac{1}{v ( v - 1 )} d v

Take the partial fraction of

\frac{1}{v ( v - 1 )} : - \frac{1}{v} + \frac{1}{v - 1}

= \frac{1}{a} \cdot \int - \frac{1}{v} + \frac{1}{v - 1} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{a} (- \int \frac{1}{v} d v + \int \frac{1}{v - 1} d v)

\int \frac{1}{v} d v = ln ∣ v ∣

\int \frac{1}{v - 1} d v = ln ∣ v - 1 ∣

= \frac{1}{a} (- ln ∣ v ∣ + ln ∣ v - 1 ∣)

Substitute back

= \frac{1}{a} (- ln \frac{a - b P}{a} + ln \frac{a - b P}{a} - 1)

Apply absolute rule:

\frac{a}{b} = \frac{∣ a ∣}{∣ b ∣}

\frac{a - b P}{a} = \frac{∣ a - b P ∣}{∣ a ∣}

= \frac{1}{a} (- ln (\frac{∣ a - b P ∣}{∣ a ∣}) + ln \frac{a - b P}{a} - 1)

Add a constant to the solution

= \frac{1}{a} (- ln (\frac{∣ a - b P ∣}{∣ a ∣}) + ln \frac{a - b P}{a} - 1) + C

\frac{\partial}{\partial y} (\frac{x ^{2}}{y + 1})

\frac{\partial}{\partial x} (- sin (2 x))

y = \frac{( x ^{2} + 3 )}{2 x}

\frac{d}{d x} (ln (12 cos^{2} (x)))

x \to 0 lim (x lo g_{e} (x))

What is the integral of 1/(P(a-bP)) ?
The integral of 1/(P(a-bP)) is 1/a (-ln((|a-bP|}{|a|})+ln|\frac{a-bP)/a-1|)+C