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

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

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

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

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

Solution steps

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

Simplify

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

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

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