What is the integral of (sqrt(81-x^2))/(x^2) ?

The integral of (sqrt(81-x^2))/(x^2) is -arcsin(1/9 x)-(sqrt(81-x^2))/x+C

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integral of (sqrt(81-x^2))/(x^2)

\int \frac{81 - x ^{2}}{x ^{2}} d x

- arcsin (\frac{1}{9} x) - \frac{81 - x ^{2}}{x} + C

Solution steps

\int \frac{81 - x ^{2}}{x ^{2}} d x

Apply Trigonometric Substitution

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

= \int cot^{2} (u) d u

Rewrite using trig identities

= \int - 1 + csc^{2} (u) d u

Apply the Sum Rule:

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

= - \int 1 d u + \int csc^{2} (u) d u

\int 1 d u = u

\int csc^{2} (u) d u = - cot (u)

= - u - cot (u)

Substitute back

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

= - arcsin (\frac{1}{9} x) - cot (arcsin (\frac{1}{9} x))

Simplify

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

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

Add a constant to the solution

= - arcsin (\frac{1}{9} x) - \frac{81 - x ^{2}}{x} + C

y = 3 x^{4} - 12 x^{3}

x \to - 1 lim (x)

f (x) = \frac{5 x ^{\frac{3}{2}} - 1 5 ^{\frac{1}{2}}}{2}

\int_{- 5}^{5} 25 - x^{2} d x

\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{3}{x ^{4}}) \cdot (x + 5 x^{3}))

What is the integral of (sqrt(81-x^2))/(x^2) ?
The integral of (sqrt(81-x^2))/(x^2) is -arcsin(1/9 x)-(sqrt(81-x^2))/x+C