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

The integral of (sqrt(x^2-9))/(x^4) is ((x^2-9)^{3/2})/(27x^3)+C

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

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

\frac{( x ^{2} - 9 ) ^{\frac{3}{2}}}{27 x ^{3}} + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{tan ^{2} ( u )}{9 sec ^{3} ( u )} d u

= \int \frac{tan ^{2} ( u )}{9 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}{9} \cdot \int \frac{tan ^{2} ( u )}{sec ^{3} ( u )} d u

Rewrite using trig identities

= \frac{1}{9} \cdot \int sin^{2} (u) cos (u) d u

Apply u-substitution

: \int v^{2} d v

= \frac{1}{9} \cdot \int v^{2} d v

Apply the Power Rule

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

= \frac{1}{9} \cdot \frac{v ^{3}}{3}

Substitute back

= \frac{1}{9} \cdot \frac{sin ^{3} ( arcsec ( \frac{1}{3} x ) )}{3}

Simplify

\frac{1}{9} \cdot \frac{sin ^{3} ( arcsec ( \frac{1}{3} x ) )}{3} : \frac{( x ^{2} - 9 ) ^{\frac{3}{2}}}{27 x ^{3}}

= \frac{( x ^{2} - 9 ) ^{\frac{3}{2}}}{27 x ^{3}}

Add a constant to the solution

= \frac{( x ^{2} - 9 ) ^{\frac{3}{2}}}{27 x ^{3}} + C

\int x - 1 - ln (x) d x

\frac{\partial}{\partial y} (y + 3)

y = x^{2} - x, y = 2 (x + 2)

x \to \infty lim (- 4)

\int_{- π}^{π} e^{t} cos (n t) d t

What is the integral of (sqrt(x^2-9))/(x^4) ?
The integral of (sqrt(x^2-9))/(x^4) is ((x^2-9)^{3/2})/(27x^3)+C