What is the integral of (25sqrt(x^2-16))/(x^4) ?

The integral of (25sqrt(x^2-16))/(x^4) is (25(x^2-16)^{3/2})/(48x^3)+C

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integral of (25sqrt(x^2-16))/(x^4)

\int \frac{25 x ^{2} - 16}{x ^{4}} d x

\frac{25 ( x ^{2} - 16 ) ^{\frac{3}{2}}}{48 x ^{3}} + C

Solution steps

\int \frac{25 x ^{2} - 16}{x ^{4}} d x

Take the constant out:

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

= 25 \cdot \int \frac{x ^{2} - 16}{x ^{4}} d x

Apply Trigonometric Substitution

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

= 25 \cdot \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

= 25 \cdot \frac{1}{16} \cdot \int \frac{tan ^{2} ( u )}{sec ^{3} ( u )} d u

Rewrite using trig identities

= 25 \cdot \frac{1}{16} \cdot \int sin^{2} (u) cos (u) d u

Apply u-substitution

: \int v^{2} d v

= 25 \cdot \frac{1}{16} \cdot \int v^{2} d v

Apply the Power Rule

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

= 25 \cdot \frac{1}{16} \cdot \frac{v ^{3}}{3}

Substitute back

= 25 \cdot \frac{1}{16} \cdot \frac{sin ^{3} ( arcsec ( \frac{1}{4} x ) )}{3}

Simplify

25 \cdot \frac{1}{16} \cdot \frac{sin ^{3} ( arcsec ( \frac{1}{4} x ) )}{3} : \frac{25 ( x ^{2} - 16 ) ^{\frac{3}{2}}}{48 x ^{3}}

= \frac{25 ( x ^{2} - 16 ) ^{\frac{3}{2}}}{48 x ^{3}}

Add a constant to the solution

= \frac{25 ( x ^{2} - 16 ) ^{\frac{3}{2}}}{48 x ^{3}} + C

y = \frac{2}{3 x ^{4}}

e^{2 x} (2 x + 3)^{7}

\frac{\partial}{\partial t} (x)

x d y = (2 x e^{x} - y + 6 x^{2}) d x

\int 2 e^{x} d x

What is the integral of (25sqrt(x^2-16))/(x^4) ?
The integral of (25sqrt(x^2-16))/(x^4) is (25(x^2-16)^{3/2})/(48x^3)+C