What is the integral of 1/(x^2sqrt(x^2-8)) ?

The integral of 1/(x^2sqrt(x^2-8)) is (sqrt(x^2-8))/(8x)+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(x^2-8))

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

\frac{x ^{2} - 8}{8 x} + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{8 sec ( u )} d u

= \int \frac{1}{8 sec ( u )} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{8} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{8} sin (u)

Substitute back

u = arcsec (\frac{1}{2 2} x)

= \frac{1}{8} sin (arcsec (\frac{1}{2 2} x))

Simplify

\frac{1}{8} sin (arcsec (\frac{1}{2 2} x)) : \frac{x ^{2} - 8}{8 x}

= \frac{x ^{2} - 8}{8 x}

Add a constant to the solution

= \frac{x ^{2} - 8}{8 x} + C

\frac{\partial}{\partial y} (ln (x^{2} + 3 x y))

f (x) = x^{2} e^{x}

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

\int_{0}^{12} 2 π x (12 x - x^{2}) d x

(u - u) (u + u)

What is the integral of 1/(x^2sqrt(x^2-8)) ?
The integral of 1/(x^2sqrt(x^2-8)) is (sqrt(x^2-8))/(8x)+C