What is the integral of 1/(sqrt((4-x^2)^3)) ?

The integral of 1/(sqrt((4-x^2)^3)) is x/(4sqrt(4-x^2))+C

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integral of 1/(sqrt((4-x^2)^3))

\int \frac{1}{( 4 - x ^{2} ) ^{3}} d x

\frac{x}{4 4 - x ^{2}} + C

Solution steps

\int \frac{1}{( 4 - x ^{2} ) ^{3}} d x

(4 - x^{2})^{3} = (4 - x^{2})^{\frac{3}{2}},

assuming

(4 - x^{2}) \geq 0

= \int \frac{1}{( 4 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{4 cos ^{2} ( u )} d u

= \int \frac{1}{4 cos ^{2} ( u )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{cos ^{2} ( u )} d u

Use the common integral:

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

= \frac{1}{4} tan (u)

Substitute back

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

= \frac{1}{4} tan (arcsin (\frac{1}{2} x))

Simplify

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

= \frac{x}{4 4 - x ^{2}}

Add a constant to the solution

= \frac{x}{4 4 - x ^{2}} + C

y - 1 + x y^{^{'}} = 4 x

- 5 x^{2}

\frac{\partial}{\partial x} (\frac{( x ^{2} )}{y + 1})

\int \frac{y}{( y ^{2} - 1 ) ^{\frac{3}{2}}} d y

(ln (x + 1))^{^{'}}

What is the integral of 1/(sqrt((4-x^2)^3)) ?
The integral of 1/(sqrt((4-x^2)^3)) is x/(4sqrt(4-x^2))+C