What is the integral of (162)/(x^2sqrt(81-x^2)) ?

The integral of (162)/(x^2sqrt(81-x^2)) is -(2sqrt(-x^2+81))/x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (162)/(x^2sqrt(81-x^2))

\int \frac{162}{x ^{2} 81 - x ^{2}} d x

- \frac{2 - x ^{2} + 81}{x} + C

Solution steps

\int \frac{162}{x ^{2} 81 - x ^{2}} d x

Take the constant out:

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

= 162 \cdot \int \frac{1}{x ^{2} 81 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{81 cos ( u ) sin ( u )} d u

= 162 \cdot \int \frac{cot ( u )}{81 cos ( u ) sin ( u )} d u

Take the constant out:

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

= 162 \cdot \frac{1}{81} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= 162 \cdot \frac{1}{81} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= 162 \cdot \frac{1}{81} (- cot (u))

Substitute back

u = arcsin (\frac{1}{9} x)

= 162 \cdot \frac{1}{81} (- cot (arcsin (\frac{1}{9} x)))

Simplify

162 \cdot \frac{1}{81} (- cot (arcsin (\frac{1}{9} x))) : - \frac{2 - x ^{2} + 81}{x}

= - \frac{2 - x ^{2} + 81}{x}

Add a constant to the solution

= - \frac{2 - x ^{2} + 81}{x} + C

\int (cos (8 x)) d x

\frac{d}{d x} (\frac{x}{x ^{2} + 8})

n = 1 \sum \infty n \cdot x^{n}

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

x \frac{d y}{d x} + x y = 1

What is the integral of (162)/(x^2sqrt(81-x^2)) ?
The integral of (162)/(x^2sqrt(81-x^2)) is -(2sqrt(-x^2+81))/x+C