What is the integral of 1/(x^3sqrt(16-x^2)) ?

The integral of 1/(x^3sqrt(16-x^2)) is

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integral of 1/(x^3sqrt(16-x^2))

\int \frac{1}{x ^{3} 16 - x ^{2}} d x

\frac{1}{512} (- cot^{2} (\frac{arcsin ( \frac{x}{4} )}{2}) + 4 ln tan (\frac{arcsin ( \frac{x}{4} )}{2}) + tan^{2} (\frac{arcsin ( \frac{1}{4} x )}{2})) + C

Solution steps

\int \frac{1}{x ^{3} 16 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{64 sin ^{3} ( u )} d u

= \int \frac{1}{64 sin ^{3} ( u )} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int \frac{1}{sin ^{3} ( u )} d u

Apply u-substitution

: \int \frac{( 1 + v ^{2} ) ^{2}}{4 v ^{3}} d v

= \frac{1}{64} \cdot \int \frac{( 1 + v ^{2} ) ^{2}}{4 v ^{3}} d v

Take the constant out:

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

= \frac{1}{64} \cdot \frac{1}{4} \cdot \int \frac{( 1 + v ^{2} ) ^{2}}{v ^{3}} d v

Expand

\frac{( 1 + v ^{2} ) ^{2}}{v ^{3}} : \frac{1}{v ^{3}} + \frac{2}{v} + v

= \frac{1}{64} \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{3}} + \frac{2}{v} + v d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{64} \cdot \frac{1}{4} (\int \frac{1}{v ^{3}} d v + \int \frac{2}{v} d v + \int v d v)

\int \frac{1}{v ^{3}} d v = - \frac{1}{2 v ^{2}}

\int \frac{2}{v} d v = 2 ln ∣ v ∣

\int v d v = \frac{v ^{2}}{2}

= \frac{1}{64} \cdot \frac{1}{4} (- \frac{1}{2 v ^{2}} + 2 ln ∣ v ∣ + \frac{v ^{2}}{2})

Substitute back

= \frac{1}{64} \cdot \frac{1}{4} - \frac{1}{2 tan ^{2} ( \frac{a r c s i n ( \frac{1}{4} x )}{2} )} + 2 ln tan (\frac{arcsin ( \frac{1}{4} x )}{2}) + \frac{tan ^{2} ( \frac{a r c s i n ( \frac{1}{4} x )}{2} )}{2}

Simplify

\frac{1}{64} \cdot \frac{1}{4} - \frac{1}{2 tan ^{2} ( \frac{a r c s i n ( \frac{1}{4} x )}{2} )} + 2 ln tan (\frac{arcsin ( \frac{1}{4} x )}{2}) + \frac{tan ^{2} ( \frac{a r c s i n ( \frac{1}{4} x )}{2} )}{2} : \frac{1}{512} (- cot^{2} (\frac{arcsin ( \frac{x}{4} )}{2}) + 4 ln tan (\frac{arcsin ( \frac{x}{4} )}{2}) + tan^{2} (\frac{arcsin ( \frac{1}{4} x )}{2}))

= \frac{1}{512} (- cot^{2} (\frac{arcsin ( \frac{x}{4} )}{2}) + 4 ln tan (\frac{arcsin ( \frac{x}{4} )}{2}) + tan^{2} (\frac{arcsin ( \frac{1}{4} x )}{2}))

Add a constant to the solution

= \frac{1}{512} (- cot^{2} (\frac{arcsin ( \frac{x}{4} )}{2}) + 4 ln tan (\frac{arcsin ( \frac{x}{4} )}{2}) + tan^{2} (\frac{arcsin ( \frac{1}{4} x )}{2})) + C

(t - 5)^{2}

\int - π d x

n = - 4 \sum \infty (- \frac{4}{7})^{n}

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

\frac{d}{d x} (\frac{3}{1 + 3 x})

What is the integral of 1/(x^3sqrt(16-x^2)) ?
The integral of 1/(x^3sqrt(16-x^2)) is