What is the integral of (18)/((1-x^2)^{3/2)} ?

The integral of (18)/((1-x^2)^{3/2)} is (18x)/((1-x^2)^{1/2)}+C

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integral of (18)/((1-x^2)^{3/2)}

\int \frac{18}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{18 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{18}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

Take the constant out:

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

= 18 \cdot \int \frac{1}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

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

= 18 \cdot \int sec^{2} (u) d u

Use the common integral:

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

= 18 tan (u)

Substitute back

u = arcsin (x)

= 18 tan (arcsin (x))

Simplify

18 tan (arcsin (x)) : \frac{18 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{18 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{18 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} + C

n = 0 \sum \infty \frac{5 ^{n + 1}}{4 ^{n} ( 3 n - 1 ) !}

x \to 4 lim (- \frac{1}{x ^{\frac{3}{2}}})

\int e^{- x} 2 x d x

(4 x y + y^{2}) d x + (2 x^{2} + 2 x y) d y = 0

\frac{d y}{d x} = \frac{x - 1}{y - 1}

What is the integral of (18)/((1-x^2)^{3/2)} ?
The integral of (18)/((1-x^2)^{3/2)} is (18x)/((1-x^2)^{1/2)}+C