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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

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

= 16 \cdot \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

= 16 \cdot \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)

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

Use the common integral:

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

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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

y^{^{'}} + y = b

x + 2

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

\frac{d}{d x} (2 - x)

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