What is the integral of ((x^2-4)^{3/2})/(x^6) ?

The integral of ((x^2-4)^{3/2})/(x^6) is ((x^2-4)^{5/2})/(20x^5)+C

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integral of ((x^2-4)^{3/2})/(x^6)

\int \frac{( x ^{2} - 4 ) ^{\frac{3}{2}}}{x ^{6}} d x

\frac{( x ^{2} - 4 ) ^{\frac{5}{2}}}{20 x ^{5}} + C

Solution steps

\int \frac{( x ^{2} - 4 ) ^{\frac{3}{2}}}{x ^{6}} d x

Apply Trigonometric Substitution

: \int \frac{tan ^{4} ( u )}{4 sec ^{5} ( u )} d u

= \int \frac{tan ^{4} ( u )}{4 sec ^{5} ( u )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{tan ^{4} ( u )}{sec ^{5} ( u )} d u

Use the basic trigonometric identity:

\frac{1}{sec ( x )} = cos (x)

= \frac{1}{4} \cdot \int cos^{5} (u) tan^{4} (u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int v^{4} d v

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

Apply the Power Rule

: \frac{v ^{5}}{5}

= \frac{1}{4} \cdot \frac{v ^{5}}{5}

Substitute back

= \frac{1}{4} \cdot \frac{sin ^{5} ( arcsec ( \frac{1}{2} x ) )}{5}

Simplify

\frac{1}{4} \cdot \frac{sin ^{5} ( arcsec ( \frac{1}{2} x ) )}{5} : \frac{( x ^{2} - 4 ) ^{\frac{5}{2}}}{20 x ^{5}}

= \frac{( x ^{2} - 4 ) ^{\frac{5}{2}}}{20 x ^{5}}

Add a constant to the solution

= \frac{( x ^{2} - 4 ) ^{\frac{5}{2}}}{20 x ^{5}} + C

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What is the integral of ((x^2-4)^{3/2})/(x^6) ?
The integral of ((x^2-4)^{3/2})/(x^6) is ((x^2-4)^{5/2})/(20x^5)+C