What is the integral of sqrt((x^2-7)/(x^8)) ?

The integral of sqrt((x^2-7)/(x^8)) is ((x^2-7)^{3/2})/(21x^3)+C

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integral of sqrt((x^2-7)/(x^8))

\int \frac{x ^{2} - 7}{x ^{8}} d x

\frac{( x ^{2} - 7 ) ^{\frac{3}{2}}}{21 x ^{3}} + C

Solution steps

\int \frac{x ^{2} - 7}{x ^{8}} d x

Simplify

\frac{x ^{2} - 7}{x ^{8}} : \frac{x ^{2} - 7}{x ^{8}}

= \int \frac{x ^{2} - 7}{x ^{8}} d x

x^{8} = x^{4},

assuming

x \geq 0

= \int \frac{x ^{2} - 7}{x ^{4}} d x

Apply Trigonometric Substitution

: \int \frac{tan ^{2} ( u )}{7 sec ^{3} ( u )} d u

= \int \frac{tan ^{2} ( u )}{7 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}{7} \cdot \int \frac{tan ^{2} ( u )}{sec ^{3} ( u )} d u

Rewrite using trig identities

= \frac{1}{7} \cdot \int sin^{2} (u) cos (u) d u

Apply u-substitution

: \int v^{2} d v

= \frac{1}{7} \cdot \int v^{2} d v

Apply the Power Rule

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

= \frac{1}{7} \cdot \frac{v ^{3}}{3}

Substitute back

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

Simplify

\frac{1}{7} \cdot \frac{sin ^{3} ( arcsec ( \frac{1}{7} x ) )}{3} : \frac{( x ^{2} - 7 ) ^{\frac{3}{2}}}{21 x ^{3}}

= \frac{( x ^{2} - 7 ) ^{\frac{3}{2}}}{21 x ^{3}}

Add a constant to the solution

= \frac{( x ^{2} - 7 ) ^{\frac{3}{2}}}{21 x ^{3}} + C

x \to 0 + lim (x^{2} ln (x^{2}))

\int (4 x - x^{2})^{2} d x

e^{t} cos (2 t)

- 1 {\frac{15}{s ^{2} + 4 s + 13}}

y = e^{x + 1} + 1

What is the integral of sqrt((x^2-7)/(x^8)) ?
The integral of sqrt((x^2-7)/(x^8)) is ((x^2-7)^{3/2})/(21x^3)+C