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

The integral of 1/(x^4sqrt(x^2+4)) is ((x^2-2)sqrt(x^2+4))/(24x^3)+C

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integral of 1/(x^4sqrt(x^2+4))

\int \frac{1}{x ^{4} x ^{2} + 4} d x

\frac{( x ^{2} - 2 ) x ^{2} + 4}{24 x ^{3}} + C

Solution steps

\int \frac{1}{x ^{4} x ^{2} + 4} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{16 tan ^{4} ( u )} d u

= \int \frac{sec ( u )}{16 tan ^{4} ( u )} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{sec ( u )}{tan ^{4} ( u )} d u

Rewrite using trig identities

= \frac{1}{16} \cdot \int \frac{( 1 - sin ^{2} ( u ) ) cos ( u )}{sin ^{4} ( u )} d u

Apply u-substitution

: \int \frac{1 - v ^{2}}{v ^{4}} d v

= \frac{1}{16} \cdot \int \frac{1 - v ^{2}}{v ^{4}} d v

Expand

\frac{1 - v ^{2}}{v ^{4}} : \frac{1}{v ^{4}} - \frac{1}{v ^{2}}

= \frac{1}{16} \cdot \int \frac{1}{v ^{4}} - \frac{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}{16} (\int \frac{1}{v ^{4}} d v - \int \frac{1}{v ^{2}} d v)

\int \frac{1}{v ^{4}} d v = - \frac{1}{3 v ^{3}}

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

= \frac{1}{16} (- \frac{1}{3 v ^{3}} - (- \frac{1}{v}))

Substitute back

= \frac{1}{16} (- \frac{1}{3 sin ^{3} ( arctan ( \frac{1}{2} x ) )} - (- \frac{1}{sin ( arctan ( \frac{1}{2} x ) )}))

Simplify

\frac{1}{16} (- \frac{1}{3 sin ^{3} ( arctan ( \frac{1}{2} x ) )} - (- \frac{1}{sin ( arctan ( \frac{1}{2} x ) )})) : \frac{( x ^{2} - 2 ) x ^{2} + 4}{24 x ^{3}}

= \frac{( x ^{2} - 2 ) x ^{2} + 4}{24 x ^{3}}

Add a constant to the solution

= \frac{( x ^{2} - 2 ) x ^{2} + 4}{24 x ^{3}} + C

(x - 3)^{- 2}

y^{^{'}} = - y (1 + x y^{2})

x \to 2 lim (\frac{x ^{3} - 2 x ^{2}}{10 x - 20})

x \to \infty lim (\frac{i}{x})

x \to 0 + lim (x + \frac{1}{x})

What is the integral of 1/(x^4sqrt(x^2+4)) ?
The integral of 1/(x^4sqrt(x^2+4)) is ((x^2-2)sqrt(x^2+4))/(24x^3)+C