What is the integral of 1/(w^2sqrt(w^2+a^2)) ?

The integral of 1/(w^2sqrt(w^2+a^2)) is -(sqrt(a^2+w^2))/(a^2w)+C

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integral of 1/(w^2sqrt(w^2+a^2))

\int \frac{1}{w ^{2} w ^{2} + a ^{2}} d w

- \frac{a ^{2} + w ^{2}}{a ^{2} w} + C

Solution steps

\int \frac{1}{w ^{2} w ^{2} + a ^{2}} d w

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{a ^{2}} (- \frac{1}{v})

Substitute back

= \frac{1}{a ^{2}} (- \frac{1}{sin ( arctan ( \frac{1}{a} w ) )})

Simplify

\frac{1}{a ^{2}} (- \frac{1}{sin ( arctan ( \frac{1}{a} w ) )}) : - \frac{a ^{2} + w ^{2}}{a ^{2} w}

= - \frac{a ^{2} + w ^{2}}{a ^{2} w}

Add a constant to the solution

= - \frac{a ^{2} + w ^{2}}{a ^{2} w} + C

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