What is the integral of 1/(y^2sqrt(y^2+16)) ?

The integral of 1/(y^2sqrt(y^2+16)) is -(sqrt(16+y^2))/(16y)+C

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integral of 1/(y^2sqrt(y^2+16))

\int \frac{1}{y ^{2} y ^{2} + 16} d y

- \frac{16 + y ^{2}}{16 y} + C

Solution steps

\int \frac{1}{y ^{2} y ^{2} + 16} d y

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{16} (- \frac{1}{v})

Substitute back

= \frac{1}{16} (- \frac{1}{sin ( arctan ( \frac{1}{4} y ) )})

Simplify

\frac{1}{16} (- \frac{1}{sin ( arctan ( \frac{1}{4} y ) )}) : - \frac{16 + y ^{2}}{16 y}

= - \frac{16 + y ^{2}}{16 y}

Add a constant to the solution

= - \frac{16 + y ^{2}}{16 y} + C

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What is the integral of 1/(y^2sqrt(y^2+16)) ?
The integral of 1/(y^2sqrt(y^2+16)) is -(sqrt(16+y^2))/(16y)+C