What is the integral of (sqrt(25-x^2))/(x^2) ?

The integral of (sqrt(25-x^2))/(x^2) is -arcsin(1/5 x)-(sqrt(25-x^2))/x+C

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integral of (sqrt(25-x^2))/(x^2)

\int \frac{25 - x ^{2}}{x ^{2}} d x

- arcsin (\frac{1}{5} x) - \frac{25 - x ^{2}}{x} + C

Solution steps

\int \frac{25 - 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{25 - x ^{2}}{x}

= - arcsin (\frac{1}{5} x) - \frac{25 - x ^{2}}{x}

Add a constant to the solution

= - arcsin (\frac{1}{5} x) - \frac{25 - x ^{2}}{x} + C

y = \frac{x ^{10} - 5 x ^{6} - 12 x}{x ^{6}}

\frac{d y}{d x} - \frac{9}{x} y = \frac{y ^{3}}{x ^{13}}, y (1) = 1

\frac{\partial}{\partial x} (r^{2} + s^{2})

\frac{\partial}{\partial x} (e^{3 x} d y)

\frac{\partial}{\partial x} (x^{2} + 6 x y + y^{2})

What is the integral of (sqrt(25-x^2))/(x^2) ?
The integral of (sqrt(25-x^2))/(x^2) is -arcsin(1/5 x)-(sqrt(25-x^2))/x+C