What is the integral of 1/(t^2sqrt(t^2+4)) ?

The integral of 1/(t^2sqrt(t^2+4)) is -(sqrt(4+t^2))/(4t)+C

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integral of 1/(t^2sqrt(t^2+4))

\int \frac{1}{t ^{2} t ^{2} + 4} d t

- \frac{4 + t ^{2}}{4 t} + C

Solution steps

\int \frac{1}{t ^{2} t ^{2} + 4} d t

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{4 tan ^{2} ( u )} d u

= \int \frac{sec ( u )}{4 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}{4} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{4} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{4} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{4} (- \frac{1}{v})

Substitute back

= \frac{1}{4} (- \frac{1}{sin ( arctan ( \frac{1}{2} t ) )})

Simplify

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

= - \frac{4 + t ^{2}}{4 t}

Add a constant to the solution

= - \frac{4 + t ^{2}}{4 t} + C

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What is the integral of 1/(t^2sqrt(t^2+4)) ?
The integral of 1/(t^2sqrt(t^2+4)) is -(sqrt(4+t^2))/(4t)+C