What is the integral of 1/((81-x^2)^{3/2)} ?

The integral of 1/((81-x^2)^{3/2)} is x/(81(81-x^2)^{1/2)}+C

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integral of 1/((81-x^2)^{3/2)}

\int \frac{1}{( 81 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{81 ( 81 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 81 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{81 cos ^{2} ( u )} d u

= \int \frac{1}{81 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}{81} \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}{81} tan (u)

Substitute back

u = arcsin (\frac{1}{9} x)

= \frac{1}{81} tan (arcsin (\frac{1}{9} x))

Simplify

\frac{1}{81} tan (arcsin (\frac{1}{9} x)) : \frac{x}{81 ( 81 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{81 ( 81 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{81 ( 81 - x ^{2} ) ^{\frac{1}{2}}} + C

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n = 1 \sum \infty \frac{n - 1}{n}

\int 12 x^{2} + 6 x - 4 d x

\frac{\partial}{\partial x} (y cos (x y) - y sin (x y))

What is the integral of 1/((81-x^2)^{3/2)} ?
The integral of 1/((81-x^2)^{3/2)} is x/(81(81-x^2)^{1/2)}+C