What is the integral of 1/((6-x^2)^{3/2)} ?

The integral of 1/((6-x^2)^{3/2)} is x/(6(6-x^2)^{1/2)}+C

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integral of 1/((6-x^2)^{3/2)}

\int \frac{1}{( 6 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{6 ( 6 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 6 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{6 cos ^{2} ( u )} d u

= \int \frac{1}{6 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}{6} \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}{6} tan (u)

Substitute back

u = arcsin (\frac{1}{6 ^{\frac{1}{2}}} x)

= \frac{1}{6} tan (arcsin (\frac{1}{6 ^{\frac{1}{2}}} x))

Simplify

\frac{1}{6} tan (arcsin (\frac{1}{6 ^{\frac{1}{2}}} x)) : \frac{x}{6 ( 6 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{6 ( 6 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{6 ( 6 - x ^{2} ) ^{\frac{1}{2}}} + C

\frac{d}{d x} (2 x - \frac{2000}{x ^{2}})

\int 6 arctan (4 y) d y

\frac{\partial}{\partial x} (2 y^{4 x})

f (x) = sin (x) + 1, at x = - \frac{π}{2}

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

What is the integral of 1/((6-x^2)^{3/2)} ?
The integral of 1/((6-x^2)^{3/2)} is x/(6(6-x^2)^{1/2)}+C