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

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

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

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

\frac{t ^{2} - 1}{t} + C

Solution steps

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

Apply Trigonometric Substitution

: \int cos (u) d u

= \int cos (u) d u

Use the common integral:

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

= sin (u)

Substitute back

u = arcsec (t)

= sin (arcsec (t))

Use the following identity:

sin (arcsec (x)) = \frac{x ^{2} - 1}{x}

= \frac{t ^{2} - 1}{t}

Add a constant to the solution

= \frac{t ^{2} - 1}{t} + C

\frac{x ^{2} - 6 x + 7}{x ^{2} + 9}

\int_{1}^{3} \frac{3}{x} - \frac{3}{x ^{2}} d x

\int x^{3} e^{- 0.25 x} d x

\frac{d}{d x} (arcsin (2 x - 1))

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