What is the integral of 1/(t^2sqrt(t^2-16)) ?

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

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integral of 1/(t^2sqrt(t^2-16))

\int \frac{1}{t ^{2} t ^{2} - 16} d t

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

Solution steps

\int \frac{1}{t ^{2} t ^{2} - 16} d t

Apply Trigonometric Substitution

: \int \frac{1}{16 sec ( u )} d u

= \int \frac{1}{16 sec ( 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{1}{sec ( u )} d u

Rewrite using trig identities

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

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{16} sin (u)

Substitute back

u = arcsec (\frac{1}{4} t)

= \frac{1}{16} sin (arcsec (\frac{1}{4} t))

Simplify

\frac{1}{16} sin (arcsec (\frac{1}{4} t)) : \frac{t ^{2} - 16}{16 t}

= \frac{t ^{2} - 16}{16 t}

Add a constant to the solution

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

\frac{d}{d x} (e^{\frac{x}{6}})

\int \frac{1}{1 + e ^{- t}} d t

\int \frac{x ^{2}}{x ^{2} + 81} d x

\frac{y ^{^{'}} - e ^{- t} + 5}{y} = - 5, y (0) = - 4

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