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

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

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

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

- \frac{16 - x ^{2}}{16 x} + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{16 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{16 cos ( u ) sin ( u )} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

\frac{cot ( u )}{cos ( u ) sin ( u )} : csc^{2} (u)

= \frac{1}{16} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{16} (- cot (u))

Substitute back

u = arcsin (\frac{1}{4} x)

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

Simplify

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

= - \frac{16 - x ^{2}}{16 x}

Add a constant to the solution

= - \frac{16 - x ^{2}}{16 x} + C

\frac{\partial}{\partial z} (2 y^{3} z)

\frac{d}{d x} (x^{2} + x + 1)

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

\int x^{5} (cos (x^{3})) d x

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