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

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

x^{7} - 6 x + 8

\frac{1}{( e ^{m} )}

\frac{x ^{2} + x ^{4}}{4 x}

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cot^{4} (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