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

Take the constant out:

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

= - \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

\int_{0}^{2} \frac{1}{4 - x ^{2}} d x

(e^{y} + 6 x) d x + x e^{y} d y = 0

x \to \frac{1}{2} lim (4 x^{2} - 8 x + 5)

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

y = 5 x^{8} e^{x}

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