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

The integral of 1/(x^4sqrt(x^2+3)) is 1/9 (-((3+x^2)^{3/2}}{3x^3}+\frac{sqrt(3+x^2))/x)+C

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integral of 1/(x^4sqrt(x^2+3))

\int \frac{1}{x ^{4} x ^{2} + 3} d x

\frac{1}{9} - \frac{( 3 + x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}} + \frac{3 + x ^{2}}{x} + C

Solution steps

\int \frac{1}{x ^{4} x ^{2} + 3} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{9 tan ^{4} ( u )} d u

= \int \frac{sec ( u )}{9 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}{9} \cdot \int \frac{sec ( u )}{tan ^{4} ( u )} d u

Rewrite using trig identities

= \frac{1}{9} \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}{9} \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}{9} \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}{9} (\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}{9} (- \frac{1}{3 v ^{3}} - (- \frac{1}{v}))

Substitute back

= \frac{1}{9} - \frac{1}{3 sin ^{3} ( arctan ( \frac{1}{3} x ) )} - - \frac{1}{sin ( arctan ( \frac{1}{3} x ) )}

Simplify

\frac{1}{9} - \frac{1}{3 sin ^{3} ( arctan ( \frac{1}{3} x ) )} - - \frac{1}{sin ( arctan ( \frac{1}{3} x ) )} : \frac{1}{9} - \frac{( 3 + x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}} + \frac{3 + x ^{2}}{x}

= \frac{1}{9} - \frac{( 3 + x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}} + \frac{3 + x ^{2}}{x}

Add a constant to the solution

= \frac{1}{9} - \frac{( 3 + x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}} + \frac{3 + x ^{2}}{x} + C

x \to 0 lim (\frac{x ^{3}}{x ^{4}})

y = 5 x^{2}, (- 2, 20)

\frac{d}{d x} (f (x) (x^{2}))

\int \frac{5 - 50 x}{1 - 25 x ^{2}} d x

n = 0 \sum \infty 2 \cdot 6^{n}

What is the integral of 1/(x^4sqrt(x^2+3)) ?
The integral of 1/(x^4sqrt(x^2+3)) is 1/9 (-((3+x^2)^{3/2}}{3x^3}+\frac{sqrt(3+x^2))/x)+C