What is the integral of (30sqrt(x^2-49))/(x^4) ?

The integral of (30sqrt(x^2-49))/(x^4) is (10(x^2-49)^{3/2})/(49x^3)+C

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integral of (30sqrt(x^2-49))/(x^4)

\int \frac{30 x ^{2} - 49}{x ^{4}} d x

\frac{10 ( x ^{2} - 49 ) ^{\frac{3}{2}}}{49 x ^{3}} + C

Solution steps

\int \frac{30 x ^{2} - 49}{x ^{4}} d x

Take the constant out:

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

= 30 \cdot \int \frac{x ^{2} - 49}{x ^{4}} d x

Apply Trigonometric Substitution

: \int \frac{tan ^{2} ( u )}{49 sec ^{3} ( u )} d u

= 30 \cdot \int \frac{tan ^{2} ( u )}{49 sec ^{3} ( u )} d u

Take the constant out:

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

= 30 \cdot \frac{1}{49} \cdot \int \frac{tan ^{2} ( u )}{sec ^{3} ( u )} d u

Rewrite using trig identities

= 30 \cdot \frac{1}{49} \cdot \int sin^{2} (u) cos (u) d u

Apply u-substitution

: \int v^{2} d v

= 30 \cdot \frac{1}{49} \cdot \int v^{2} d v

Apply the Power Rule

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

= 30 \cdot \frac{1}{49} \cdot \frac{v ^{3}}{3}

Substitute back

= 30 \cdot \frac{1}{49} \cdot \frac{sin ^{3} ( arcsec ( \frac{1}{7} x ) )}{3}

Simplify

30 \cdot \frac{1}{49} \cdot \frac{sin ^{3} ( arcsec ( \frac{1}{7} x ) )}{3} : \frac{10 ( x ^{2} - 49 ) ^{\frac{3}{2}}}{49 x ^{3}}

= \frac{10 ( x ^{2} - 49 ) ^{\frac{3}{2}}}{49 x ^{3}}

Add a constant to the solution

= \frac{10 ( x ^{2} - 49 ) ^{\frac{3}{2}}}{49 x ^{3}} + C

\frac{\partial}{\partial x} (\frac{x}{x - 1})

x \to 0 lim (2 x^{2} cos (\frac{5}{x}))

f (x) = 5 x^{5} ln (x)

\frac{d y}{d x} = \frac{y}{x} + 5 x + 4

\frac{\partial ^{2}}{\partial y ^{2}} (x^{2} - y^{2})

What is the integral of (30sqrt(x^2-49))/(x^4) ?
The integral of (30sqrt(x^2-49))/(x^4) is (10(x^2-49)^{3/2})/(49x^3)+C