What is the integral of 1/(r(2-r^2)) ?

The integral of 1/(r(2-r^2)) is 1/4 (2ln(r)-ln|r^2-2|)+C

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integral of 1/(r(2-r^2))

\int \frac{1}{r ( 2 - r ^{2} )} d r

\frac{1}{4} (2 ln (r) - ln r^{2} - 2) + C

Solution steps

\int \frac{1}{r ( 2 - r ^{2} )} d r

Apply u-substitution

: \int \frac{1}{2 u ( 2 - u )} d u

= \int \frac{1}{2 u ( 2 - u )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ( 2 - u )} d u

Complete the square

u (2 - u) : - (u - 1)^{2} + 1

= \frac{1}{2} \cdot \int \frac{1}{- ( u - 1 ) ^{2} + 1} d u

Apply u-substitution

: \int \frac{1}{- v ^{2} + 1} d v

= \frac{1}{2} \cdot \int \frac{1}{- v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{2} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= \frac{1}{2} (\frac{ln r ^{2} - 1 + 1}{2} - \frac{ln r ^{2} - 1 - 1}{2})

Simplify

\frac{1}{2} (\frac{ln r ^{2} - 1 + 1}{2} - \frac{ln r ^{2} - 1 - 1}{2}) : \frac{1}{4} (2 ln (r) - ln r^{2} - 2)

= \frac{1}{4} (2 ln (r) - ln r^{2} - 2)

Add a constant to the solution

= \frac{1}{4} (2 ln (r) - ln r^{2} - 2) + C

x \to 1 lim (\frac{3 x}{x ^{2} - 1})

e^{- 5 x}

\int_{1}^{9} \frac{1}{2 x} d x

\frac{d}{d x} (7 (2 x e^{x} + e^{x} x^{2}))

\int y + x e^{y} d y

What is the integral of 1/(r(2-r^2)) ?
The integral of 1/(r(2-r^2)) is 1/4 (2ln(r)-ln|r^2-2|)+C