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

The integral of (sqrt(16-x^2))/(x^2) is -arcsin(1/4 x)-(sqrt(16-x^2))/x+C

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

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

- arcsin (\frac{1}{4} x) - \frac{16 - x ^{2}}{x} + C

Solution steps

\int \frac{16 - 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}{4} x)

= - arcsin (\frac{1}{4} x) - cot (arcsin (\frac{1}{4} x))

Simplify

- arcsin (\frac{1}{4} x) - cot (arcsin (\frac{1}{4} x)) : - arcsin (\frac{1}{4} x) - \frac{16 - x ^{2}}{x}

= - arcsin (\frac{1}{4} x) - \frac{16 - x ^{2}}{x}

Add a constant to the solution

= - arcsin (\frac{1}{4} x) - \frac{16 - x ^{2}}{x} + C

\int \frac{1}{8 x - x ^{2}} d x

5 3 x

\frac{\partial}{\partial y} (x ln (yz))

\int x^{4} + 10 x^{2} + 25 d x

y^{^{'}} + \frac{1}{2} y = 17 sin (2 t)

What is the integral of (sqrt(16-x^2))/(x^2) ?
The integral of (sqrt(16-x^2))/(x^2) is -arcsin(1/4 x)-(sqrt(16-x^2))/x+C