What is the integral of y/((4+y^2)) ?

The integral of y/((4+y^2)) is 1/2 ln|4+y^2|+C

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integral of y/((4+y^2))

\int \frac{y}{( 4 + y ^{2} )} d y

\frac{1}{2} ln 4 + y^{2} + C

Solution steps

\int \frac{y}{4 + 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 = 4 + y^{2}

= \frac{1}{2} ln 4 + y^{2}

Add a constant to the solution

= \frac{1}{2} ln 4 + y^{2} + C

\frac{d}{d x} (lo g_{3} ((x^{2} + 2 x)^{4}))

\int x 2^{x + 2} d x

\frac{d ^{2} y}{d x ^{2}} - 9 \frac{d y}{d x} + 4 y = x e^{x}

sin (x) + \frac{3}{5} cot (x)

\frac{1}{s + 6}

What is the integral of y/((4+y^2)) ?
The integral of y/((4+y^2)) is 1/2 ln|4+y^2|+C