What is the integral of y/(y^2+9) ?

The integral of y/(y^2+9) is 1/2 ln|y^2+9|+C

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integral of y/(y^2+9)

\int \frac{y}{y ^{2} + 9} d y

\frac{1}{2} ln y^{2} + 9 + C

Solution steps

\int \frac{y}{y ^{2} + 9} 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 = y^{2} + 9

= \frac{1}{2} ln y^{2} + 9

Add a constant to the solution

= \frac{1}{2} ln y^{2} + 9 + C

e^{7}

\int \frac{24 x}{4 4 x ^{2} + 8} d x

\int 10 cos (x) d x

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 8)^{8})

\frac{d}{d x} ((x - 1) ln ∣ x - 1 ∣)

What is the integral of y/(y^2+9) ?
The integral of y/(y^2+9) is 1/2 ln|y^2+9|+C