What is the integral of y/(x(2-x^2)) ?

The integral of y/(x(2-x^2)) is y/4 (2ln(x)-ln|x^2-2|)+C

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integral of y/(x(2-x^2))

\int \frac{y}{x ( 2 - x ^{2} )} d x

\frac{y}{4} (2 ln (x) - ln x^{2} - 2) + C

Solution steps

\int \frac{y}{x ( 2 - x ^{2} )} d x

Take the constant out:

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

= y \cdot \int \frac{1}{( 2 - x ^{2} ) x} d x

Apply u-substitution

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

= y \cdot \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

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

Complete the square

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

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

Apply u-substitution

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

= y \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}

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

Substitute back

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

Simplify

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

= \frac{y}{4} (2 ln (x) - ln x^{2} - 2)

Add a constant to the solution

= \frac{y}{4} (2 ln (x) - ln x^{2} - 2) + C

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