What is the integral of (sqrt(5-3y^2))/y ?

The integral of (sqrt(5-3y^2))/y is

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integral of (sqrt(5-3y^2))/y

\int \frac{5 - 3 y ^{2}}{y} d y

- 5 (\frac{1}{2} ln \frac{5 - 3 y ^{2}}{5} + 1 - \frac{1}{2} ln \frac{5 - 3 y ^{2}}{5} - 1 - \frac{1}{5} (5 - 3 y^{2})) + C

Solution steps

\int \frac{5 - 3 y ^{2}}{y} d y

Apply u-substitution

: \int \frac{u ^{2}}{u ^{2} - 5} d u

= \int \frac{u ^{2}}{u ^{2} - 5} d u

Factor

u^{2} - 5 : - (- u^{2} + 5)

= \int \frac{u ^{2}}{- ( - u ^{2} + 5 )} d u

Take the constant out:

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

= - \int \frac{u ^{2}}{- u ^{2} + 5} d u

Apply Integral Substitution

= - \int \frac{5 v ^{2}}{- v ^{2} + 1} d v

Take the constant out:

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

= - 5 \cdot \int \frac{v ^{2}}{- v ^{2} + 1} d v

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

= - 5 \cdot \int \frac{1}{- v ^{2} + 1} - 1 d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - 5 (\int \frac{1}{- v ^{2} + 1} d v - \int 1 d v)

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

\int 1 d v = v

= - 5 (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2} - v)

Substitute back

= - 5 \frac{ln \frac{5 - 3 y ^{2}}{5} + 1}{2} - \frac{ln \frac{5 - 3 y ^{2}}{5} - 1}{2} - \frac{5 - 3 y ^{2}}{5}

Simplify

- 5 \frac{ln \frac{5 - 3 y ^{2}}{5} + 1}{2} - \frac{ln \frac{5 - 3 y ^{2}}{5} - 1}{2} - \frac{5 - 3 y ^{2}}{5} : - 5 (\frac{1}{2} ln \frac{5 - 3 y ^{2}}{5} + 1 - \frac{1}{2} ln \frac{5 - 3 y ^{2}}{5} - 1 - \frac{1}{5} (5 - 3 y^{2}))

= - 5 (\frac{1}{2} ln \frac{5 - 3 y ^{2}}{5} + 1 - \frac{1}{2} ln \frac{5 - 3 y ^{2}}{5} - 1 - \frac{1}{5} (5 - 3 y^{2}))

Add a constant to the solution

= - 5 (\frac{1}{2} ln \frac{5 - 3 y ^{2}}{5} + 1 - \frac{1}{2} ln \frac{5 - 3 y ^{2}}{5} - 1 - \frac{1}{5} (5 - 3 y^{2})) + C

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