What is the integral of (2y)/(y^2-5) ?

The integral of (2y)/(y^2-5) is ln|y^2-5|+C

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integral of (2y)/(y^2-5)

\int \frac{2 y}{y ^{2} - 5} d y

ln y^{2} - 5 + C

Solution steps

\int \frac{2 y}{y ^{2} - 5} d y

Take the constant out:

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

= 2 \cdot \int \frac{y}{y ^{2} - 5} d y

Apply u-substitution

: \int \frac{1}{2 u} d u

= 2 \cdot \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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = y^{2} - 5

= 2 \cdot \frac{1}{2} ln y^{2} - 5

Simplify

2 \cdot \frac{1}{2} ln y^{2} - 5 : ln y^{2} - 5

= ln y^{2} - 5

Add a constant to the solution

= ln y^{2} - 5 + C

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What is the integral of (2y)/(y^2-5) ?
The integral of (2y)/(y^2-5) is ln|y^2-5|+C