What is the integral of 1/(x^4sqrt(16x^2-9)) ?

The integral of 1/(x^4sqrt(16x^2-9)) is 64/81 ((sqrt(16x^2-9))/(4x)-((16x^2-9)^{3/2})/(192x^3))+C

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integral of 1/(x^4sqrt(16x^2-9))

\int \frac{1}{x ^{4} 16 x ^{2} - 9} d x

\frac{64}{81} \frac{16 x ^{2} - 9}{4 x} - \frac{( 16 x ^{2} - 9 ) ^{\frac{3}{2}}}{192 x ^{3}} + C

Solution steps

\int \frac{1}{x ^{4} 16 x ^{2} - 9} d x

Apply Trigonometric Substitution

= \int \frac{64}{81 sec ^{3} ( u )} d u

Take the constant out:

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

= \frac{64}{81} \cdot \int \frac{1}{sec ^{3} ( u )} d u

Rewrite using trig identities

= \frac{64}{81} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

= \frac{64}{81} \cdot \int 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{64}{81} (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= \frac{64}{81} (v - \frac{v ^{3}}{3})

Substitute back

= \frac{64}{81} (sin (arcsec (\frac{4}{3} x)) - \frac{sin ^{3} ( arcsec ( \frac{4}{3} x ) )}{3})

Simplify

\frac{64}{81} (sin (arcsec (\frac{4}{3} x)) - \frac{sin ^{3} ( arcsec ( \frac{4}{3} x ) )}{3}) : \frac{64}{81} \frac{16 x ^{2} - 9}{4 x} - \frac{( 16 x ^{2} - 9 ) ^{\frac{3}{2}}}{192 x ^{3}}

= \frac{64}{81} \frac{16 x ^{2} - 9}{4 x} - \frac{( 16 x ^{2} - 9 ) ^{\frac{3}{2}}}{192 x ^{3}}

Add a constant to the solution

= \frac{64}{81} \frac{16 x ^{2} - 9}{4 x} - \frac{( 16 x ^{2} - 9 ) ^{\frac{3}{2}}}{192 x ^{3}} + C

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What is the integral of 1/(x^4sqrt(16x^2-9)) ?
The integral of 1/(x^4sqrt(16x^2-9)) is 64/81 ((sqrt(16x^2-9))/(4x)-((16x^2-9)^{3/2})/(192x^3))+C