What is the integral of 1/(y^4sqrt(y^2-5)) ?

The integral of 1/(y^4sqrt(y^2-5)) is 1/25 ((sqrt(y^2-5))/y-((y^2-5)^{3/2})/(3y^3))+C

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integral of 1/(y^4sqrt(y^2-5))

\int \frac{1}{y ^{4} y ^{2} - 5} d y

\frac{1}{25} \frac{y ^{2} - 5}{y} - \frac{( y ^{2} - 5 ) ^{\frac{3}{2}}}{3 y ^{3}} + C

Solution steps

\int \frac{1}{y ^{4} y ^{2} - 5} d y

Apply Trigonometric Substitution

: \int \frac{1}{25 sec ^{3} ( u )} d u

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

Rewrite using trig identities

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

Apply u-substitution

: \int 1 - v^{2} d v

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{25} (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= \frac{1}{25} (v - \frac{v ^{3}}{3})

Substitute back

= \frac{1}{25} sin (arcsec (\frac{1}{5} y)) - \frac{sin ^{3} ( arcsec ( \frac{1}{5} y ) )}{3}

Simplify

\frac{1}{25} sin (arcsec (\frac{1}{5} y)) - \frac{sin ^{3} ( arcsec ( \frac{1}{5} y ) )}{3} : \frac{1}{25} \frac{y ^{2} - 5}{y} - \frac{( y ^{2} - 5 ) ^{\frac{3}{2}}}{3 y ^{3}}

= \frac{1}{25} \frac{y ^{2} - 5}{y} - \frac{( y ^{2} - 5 ) ^{\frac{3}{2}}}{3 y ^{3}}

Add a constant to the solution

= \frac{1}{25} \frac{y ^{2} - 5}{y} - \frac{( y ^{2} - 5 ) ^{\frac{3}{2}}}{3 y ^{3}} + C

ln (k x)

y^{^{'}} - 3 y = 2 x + 1

\int \frac{x}{1 + 6 x - 3 x ^{2}} d x

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

x y^{^{'}} + y = y^{- 2}

What is the integral of 1/(y^4sqrt(y^2-5)) ?
The integral of 1/(y^4sqrt(y^2-5)) is 1/25 ((sqrt(y^2-5))/y-((y^2-5)^{3/2})/(3y^3))+C