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

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

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

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

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

Solution steps

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

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 (x)

= sin (arcsec (x))

Use the following identity:

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

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

Add a constant to the solution

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

f (x) = cos (8 x^{2} - 16) sin (x^{2})

(\frac{3}{5} x)^{5}

\frac{1}{9 x}

\int \frac{37 x + 5}{( 6 x + 1 ) ( x - 1 )} d x

f (x) = x \cdot ln (x)

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