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

The integral of (5s)/(sqrt(3-2s^2)) is -5/2 sqrt(3-2s^2)+C

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

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

- \frac{5}{2} 3 - 2 s^{2} + C

Solution steps

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

Take the constant out:

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

= 5 \cdot \int \frac{s}{3 - 2 s ^{2}} d s

Apply u-substitution

: \int - \frac{1}{4 u} d u

= 5 \cdot \int - \frac{1}{4 u} d u

Take the constant out:

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

= 5 (- \frac{1}{4} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= 5 (- \frac{1}{4} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 3 - 2 s^{2}

= 5 (- \frac{1}{4} \cdot 2 (3 - 2 s^{2})^{\frac{1}{2}})

Simplify

5 (- \frac{1}{4} \cdot 2 (3 - 2 s^{2})^{\frac{1}{2}}) : - \frac{5}{2} 3 - 2 s^{2}

= - \frac{5}{2} 3 - 2 s^{2}

Add a constant to the solution

= - \frac{5}{2} 3 - 2 s^{2} + C

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What is the integral of (5s)/(sqrt(3-2s^2)) ?
The integral of (5s)/(sqrt(3-2s^2)) is -5/2 sqrt(3-2s^2)+C