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

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

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

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

- ln 1 - y^{2} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- \frac{1}{2} ln ∣ u ∣)

Substitute back

u = 1 - y^{2}

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

Simplify

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

= - ln 1 - y^{2}

Add a constant to the solution

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

g (x) = x e^{x}

\int \frac{4}{x ^{2}} - \frac{7}{x ^{5}} d x

f (x) = x^{3} + 7 x - 7, (1, 1)

y^{^{''''}} = 0

y^{^{''}} - 6 y^{^{'}} + 25 y = 64 e^{- t}

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