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

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

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

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

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

Solution steps

\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{1}{2}}}

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

Add a constant to the solution

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

\frac{d}{d x} (- x^{3} + 5 a x^{2} - 3 a^{2} x - b)

\int \frac{1}{10} x d x

e^{t} cos (3 t)

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

y^{^{''}} - 10 y^{^{'}} + 2 y = 0

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