What is the integral of 1/(x^4sqrt(x^2-64)) ?

The integral of 1/(x^4sqrt(x^2-64)) is ((x^2+32)sqrt(x^2-64))/(6144x^3)+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^4sqrt(x^2-64))

\int \frac{1}{x ^{4} x ^{2} - 64} d x

\frac{( x ^{2} + 32 ) x ^{2} - 64}{6144 x ^{3}} + C

Solution steps

\int \frac{1}{x ^{4} x ^{2} - 64} d x

Apply Trigonometric Substitution

: \int \frac{1}{4096 sec ^{3} ( u )} d u

= \int \frac{1}{4096 sec ^{3} ( u )} d u

Take the constant out:

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

= \frac{1}{4096} \cdot \int \frac{1}{sec ^{3} ( u )} d u

Rewrite using trig identities

= \frac{1}{4096} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

: \int 1 - v^{2} d v

= \frac{1}{4096} \cdot \int 1 - v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4096} (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= \frac{1}{4096} (v - \frac{v ^{3}}{3})

Substitute back

= \frac{1}{4096} (sin (arcsec (\frac{1}{8} x)) - \frac{sin ^{3} ( arcsec ( \frac{1}{8} x ) )}{3})

Simplify

\frac{1}{4096} (sin (arcsec (\frac{1}{8} x)) - \frac{sin ^{3} ( arcsec ( \frac{1}{8} x ) )}{3}) : \frac{( x ^{2} + 32 ) x ^{2} - 64}{6144 x ^{3}}

= \frac{( x ^{2} + 32 ) x ^{2} - 64}{6144 x ^{3}}

Add a constant to the solution

= \frac{( x ^{2} + 32 ) x ^{2} - 64}{6144 x ^{3}} + C

\frac{\partial}{\partial x} (x^{l n (x)})

\int_{- 2}^{2} 2 x^{2} - 6 x + 14 d x

\int x^{2} - 2 x + 8 d x

\frac{d}{d x} (\frac{2 x ^{2} + 6 x + 4}{x})

x \to \infty lim (\frac{n !}{n})

What is the integral of 1/(x^4sqrt(x^2-64)) ?
The integral of 1/(x^4sqrt(x^2-64)) is ((x^2+32)sqrt(x^2-64))/(6144x^3)+C