What is the integral of (v^2)/(sqrt(-v^2+3)) ?

The integral of (v^2)/(sqrt(-v^2+3)) is 3/2 (arcsin(1/(sqrt(3))v)-1/2 sin(2arcsin(1/(sqrt(3))v)))+C

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integral of (v^2)/(sqrt(-v^2+3))

\int \frac{v ^{2}}{- v ^{2} + 3} d v

\frac{3}{2} (arcsin (\frac{1}{3} v) - \frac{1}{2} sin (2 arcsin (\frac{1}{3} v))) + C

Solution steps

\int \frac{v ^{2}}{- v ^{2} + 3} d v

Apply Trigonometric Substitution

: \int 3 sin^{2} (u) d u

= \int 3 sin^{2} (u) d u

Take the constant out:

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

= 3 \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= 3 \cdot \int \frac{1 - cos ( 2 u )}{2} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int 1 - cos (2 u) d u

Apply the Sum Rule:

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

= 3 \cdot \frac{1}{2} (\int 1 d u - \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= 3 \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{3} v)

= 3 \cdot \frac{1}{2} (arcsin (\frac{1}{3} v) - \frac{1}{2} sin (2 arcsin (\frac{1}{3} v)))

Simplify

3 \cdot \frac{1}{2} (arcsin (\frac{1}{3} v) - \frac{1}{2} sin (2 arcsin (\frac{1}{3} v))) : \frac{3}{2} (arcsin (\frac{1}{3} v) - \frac{1}{2} sin (2 arcsin (\frac{1}{3} v)))

= \frac{3}{2} (arcsin (\frac{1}{3} v) - \frac{1}{2} sin (2 arcsin (\frac{1}{3} v)))

Add a constant to the solution

= \frac{3}{2} (arcsin (\frac{1}{3} v) - \frac{1}{2} sin (2 arcsin (\frac{1}{3} v))) + C

\int sin (3 x + 4) d x

\frac{d}{d x} (\frac{2}{( x - 3 t ) ^{2} + 1})

x \to 1 lim (x (x - 2))

x \to 0 + lim (x)

n = 0 \sum \infty sin (\frac{nπ}{7})

What is the integral of (v^2)/(sqrt(-v^2+3)) ?
The integral of (v^2)/(sqrt(-v^2+3)) is 3/2 (arcsin(1/(sqrt(3))v)-1/2 sin(2arcsin(1/(sqrt(3))v)))+C