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))

integral

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

- \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

\frac{1}{x} (arctan (\frac{1}{x}))

- 4 e^{3} x

tan (\frac{x}{2})

(2 x - x^{2} + 7) e^{2} x

ln (x^{2} + 1) + 6

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