What is the integral of (y^2)/((4-y^2)^{3/2)} ?

The integral of (y^2)/((4-y^2)^{3/2)} is -arcsin(1/2 y)+y/((4-y^2)^{1/2)}+C

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integral of (y^2)/((4-y^2)^{3/2)}

\int \frac{y ^{2}}{( 4 - y ^{2} ) ^{\frac{3}{2}}} d y

- arcsin (\frac{1}{2} y) + \frac{y}{( 4 - y ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{y ^{2}}{( 4 - y ^{2} ) ^{\frac{3}{2}}} d y

Apply Trigonometric Substitution

: \int tan^{2} (u) d u

= \int tan^{2} (u) d u

Rewrite using trig identities

= \int - 1 + sec^{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

= - \int 1 d u + \int sec^{2} (u) d u

\int 1 d u = u

\int sec^{2} (u) d u = tan (u)

= - u + tan (u)

Substitute back

u = arcsin (\frac{1}{2} y)

= - arcsin (\frac{1}{2} y) + tan (arcsin (\frac{1}{2} y))

Simplify

- arcsin (\frac{1}{2} y) + tan (arcsin (\frac{1}{2} y)) : - arcsin (\frac{1}{2} y) + \frac{y}{( 4 - y ^{2} ) ^{\frac{1}{2}}}

= - arcsin (\frac{1}{2} y) + \frac{y}{( 4 - y ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= - arcsin (\frac{1}{2} y) + \frac{y}{( 4 - y ^{2} ) ^{\frac{1}{2}}} + C

arctan (\frac{x}{3})

\frac{d y}{d x} = - 6 x y

f (x) = - 18 x - \frac{17}{x}

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

\frac{d}{d x} (e^{- y x})

What is the integral of (y^2)/((4-y^2)^{3/2)} ?
The integral of (y^2)/((4-y^2)^{3/2)} is -arcsin(1/2 y)+y/((4-y^2)^{1/2)}+C