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

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

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integral of y/(1-y^2)

\int \frac{y}{1 - y ^{2}} d y

- \frac{1}{2} ln 1 - y^{2} + C

Solution steps

\int \frac{y}{1 - y ^{2}} d y

Apply u-substitution

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

= \int - \frac{1}{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} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 - y^{2}

= - \frac{1}{2} ln 1 - y^{2}

Add a constant to the solution

= - \frac{1}{2} ln 1 - y^{2} + C

\frac{\partial}{\partial x} (- 1 + ln (x))

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

\frac{d}{d x} (\frac{3 x ^{2} - 5}{2 x ^{2} + 1})

x \to - 8 + lim (\frac{( x + 7 )}{x + 8})

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