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

The integral of (sqrt(x^2-16))/(x^4) is ((x^2-16)^{3/2})/(48x^3)+C

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

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

\frac{( x ^{2} - 16 ) ^{\frac{3}{2}}}{48 x ^{3}} + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{tan ^{2} ( u )}{16 sec ^{3} ( u )} d u

= \int \frac{tan ^{2} ( u )}{16 sec ^{3} ( 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{tan ^{2} ( u )}{sec ^{3} ( u )} d u

Rewrite using trig identities

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

Apply u-substitution

: \int v^{2} d v

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

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{1}{16} \cdot \frac{v ^{3}}{3}

Substitute back

= \frac{1}{16} \cdot \frac{sin ^{3} ( arcsec ( \frac{1}{4} x ) )}{3}

Simplify

\frac{1}{16} \cdot \frac{sin ^{3} ( arcsec ( \frac{1}{4} x ) )}{3} : \frac{( x ^{2} - 16 ) ^{\frac{3}{2}}}{48 x ^{3}}

= \frac{( x ^{2} - 16 ) ^{\frac{3}{2}}}{48 x ^{3}}

Add a constant to the solution

= \frac{( x ^{2} - 16 ) ^{\frac{3}{2}}}{48 x ^{3}} + C

n = 1 \sum \infty \frac{7}{n ( n + 1 )}

x \to 1 + lim (\frac{1}{x ^{1 - x}})

f (x) = 3 (x + 1)

\frac{d y}{d x} - y = x e^{x}

x \to 3 lim (3 x^{2} - 7 x + 2)

What is the integral of (sqrt(x^2-16))/(x^4) ?
The integral of (sqrt(x^2-16))/(x^4) is ((x^2-16)^{3/2})/(48x^3)+C