What is the integral of 1/(x^2sqrt(169-x^2)) ?

The integral of 1/(x^2sqrt(169-x^2)) is -(sqrt(169-x^2))/(169x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^2sqrt(169-x^2))

\int \frac{1}{x ^{2} 169 - x ^{2}} d x

- \frac{169 - x ^{2}}{169 x} + C

Solution steps

\int \frac{1}{x ^{2} 169 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{169 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{169 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}{169} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= \frac{1}{169} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{169} (- cot (u))

Substitute back

u = arcsin (\frac{1}{13} x)

= \frac{1}{169} (- cot (arcsin (\frac{1}{13} x)))

Simplify

\frac{1}{169} (- cot (arcsin (\frac{1}{13} x))) : - \frac{169 - x ^{2}}{169 x}

= - \frac{169 - x ^{2}}{169 x}

Add a constant to the solution

= - \frac{169 - x ^{2}}{169 x} + C

\int \frac{ln ( x )}{3 x} d x

f (x) = \frac{x}{x ^{2} + 121}

\frac{d}{d x} (x^{3} (3 x + 1))

\frac{d}{d x} (\frac{1}{x ^{2} - x - 6})

x \frac{d y}{d x} = y + x e^{\frac{y}{x}}

What is the integral of 1/(x^2sqrt(169-x^2)) ?
The integral of 1/(x^2sqrt(169-x^2)) is -(sqrt(169-x^2))/(169x)+C