What is the integral of 1/(x^4sqrt(x^2-5)) ?

The integral of 1/(x^4sqrt(x^2-5)) is 1/25 ((sqrt(x^2-5))/x-((x^2-5)^{3/2})/(3x^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^2-5))

\int \frac{1}{x ^{4} x ^{2} - 5} d x

\frac{1}{25} \frac{x ^{2} - 5}{x} - \frac{( x ^{2} - 5 ) ^{\frac{3}{2}}}{3 x ^{3}} + C

Solution steps

\int \frac{1}{x ^{4} x ^{2} - 5} d x

Apply Trigonometric Substitution

: \int \frac{1}{25 sec ^{3} ( u )} d u

= \int \frac{1}{25 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}{25} \cdot \int \frac{1}{sec ^{3} ( u )} d u

Rewrite using trig identities

= \frac{1}{25} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

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

= \frac{1}{25} \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}{25} (\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}{25} (v - \frac{v ^{3}}{3})

Substitute back

= \frac{1}{25} sin (arcsec (\frac{1}{5} x)) - \frac{sin ^{3} ( arcsec ( \frac{1}{5} x ) )}{3}

Simplify

\frac{1}{25} sin (arcsec (\frac{1}{5} x)) - \frac{sin ^{3} ( arcsec ( \frac{1}{5} x ) )}{3} : \frac{1}{25} \frac{x ^{2} - 5}{x} - \frac{( x ^{2} - 5 ) ^{\frac{3}{2}}}{3 x ^{3}}

= \frac{1}{25} \frac{x ^{2} - 5}{x} - \frac{( x ^{2} - 5 ) ^{\frac{3}{2}}}{3 x ^{3}}

Add a constant to the solution

= \frac{1}{25} \frac{x ^{2} - 5}{x} - \frac{( x ^{2} - 5 ) ^{\frac{3}{2}}}{3 x ^{3}} + C

\frac{\partial}{\partial x} (cos (x + y + z))

k = 1 \sum \infty \frac{2}{3} (1.5)^{k - 1}

f (x) = cos (3 x^{2})

\frac{x}{x + 7}

y^{^{'}} = \frac{3 x ^{2} 16 + y ^{2}}{y}

What is the integral of 1/(x^4sqrt(x^2-5)) ?
The integral of 1/(x^4sqrt(x^2-5)) is 1/25 ((sqrt(x^2-5))/x-((x^2-5)^{3/2})/(3x^3))+C