What is the integral of ((4+x^2)^{1/2})/(x^6) ?

The integral of ((4+x^2)^{1/2})/(x^6) is ((x^2-6)(x^2+4)^{3/2})/(120x^5)+C

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integral of ((4+x^2)^{1/2})/(x^6)

\int \frac{( 4 + x ^{2} ) ^{\frac{1}{2}}}{x ^{6}} d x

\frac{( x ^{2} - 6 ) ( x ^{2} + 4 ) ^{\frac{3}{2}}}{120 x ^{5}} + C

Solution steps

\int \frac{( 4 + x ^{2} ) ^{\frac{1}{2}}}{x ^{6}} d x

Apply Trigonometric Substitution

= \int \frac{sec ^{3} ( u )}{16 tan ^{6} ( 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 ^{3} ( u )}{tan ^{6} ( u )} d u

Use the basic trigonometric identity:

\frac{1}{tan ( x )} = cot (x)

= \frac{1}{16} \cdot \int cot^{6} (u) sec^{3} (u) d u

Rewrite using trig identities

= \frac{1}{16} \cdot \int \frac{( 1 - sin ^{2} ( u ) ) cos ( u )}{sin ^{6} ( u )} d u

Apply u-substitution

= \frac{1}{16} \cdot \int \frac{1 - v ^{2}}{v ^{6}} d v

Expand

\frac{1 - v ^{2}}{v ^{6}} : \frac{1}{v ^{6}} - \frac{1}{v ^{4}}

= \frac{1}{16} \cdot \int \frac{1}{v ^{6}} - \frac{1}{v ^{4}} 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 ^{6}} d v - \int \frac{1}{v ^{4}} d v)

\int \frac{1}{v ^{6}} d v = - \frac{1}{5 v ^{5}}

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

= \frac{1}{16} (- \frac{1}{5 v ^{5}} - (- \frac{1}{3 v ^{3}}))

Substitute back

= \frac{1}{16} (- \frac{1}{5 sin ^{5} ( arctan ( \frac{1}{2} x ) )} - (- \frac{1}{3 sin ^{3} ( arctan ( \frac{1}{2} x ) )}))

Simplify

\frac{1}{16} (- \frac{1}{5 sin ^{5} ( arctan ( \frac{1}{2} x ) )} - (- \frac{1}{3 sin ^{3} ( arctan ( \frac{1}{2} x ) )})) : \frac{( x ^{2} - 6 ) ( x ^{2} + 4 ) ^{\frac{3}{2}}}{120 x ^{5}}

= \frac{( x ^{2} - 6 ) ( x ^{2} + 4 ) ^{\frac{3}{2}}}{120 x ^{5}}

Add a constant to the solution

= \frac{( x ^{2} - 6 ) ( x ^{2} + 4 ) ^{\frac{3}{2}}}{120 x ^{5}} + C

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What is the integral of ((4+x^2)^{1/2})/(x^6) ?
The integral of ((4+x^2)^{1/2})/(x^6) is ((x^2-6)(x^2+4)^{3/2})/(120x^5)+C