What is the integral of (sqrt(5-x^2))/(x^2) ?

The integral of (sqrt(5-x^2))/(x^2) is -arcsin(1/(sqrt(5))x)-(sqrt(5-x^2))/x+C

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integral of (sqrt(5-x^2))/(x^2)

\int \frac{5 - x ^{2}}{x ^{2}} d x

- arcsin (\frac{1}{5} x) - \frac{5 - x ^{2}}{x} + C

Solution steps

\int \frac{5 - 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}{5} x)

= - arcsin (\frac{1}{5} x) - cot (arcsin (\frac{1}{5} x))

Simplify

- arcsin (\frac{1}{5} x) - cot (arcsin (\frac{1}{5} x)) : - arcsin (\frac{1}{5} x) - \frac{5 - x ^{2}}{x}

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

Add a constant to the solution

= - arcsin (\frac{1}{5} x) - \frac{5 - x ^{2}}{x} + C

\int 1 + tan^{2} (x) - 2 cos (x) d x

\frac{\partial}{\partial r} (\frac{π}{3} h^{2} (3 r - h))

e^{x} y d y = (e^{- y} + e^{2 x - y}) d x

n = 1 \sum \infty n e^{5 n}

\int_{0}^{2} 10000 e^{0.04 x} d x

What is the integral of (sqrt(5-x^2))/(x^2) ?
The integral of (sqrt(5-x^2))/(x^2) is -arcsin(1/(sqrt(5))x)-(sqrt(5-x^2))/x+C