What is the integral of 1/(x^2*sqrt(x^2-4)) ?

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

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integral of 1/(x^2*sqrt(x^2-4))

\int \frac{1}{x ^{2} \cdot x ^{2} - 4} d x

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

Solution steps

\int \frac{1}{x ^{2} x ^{2} - 4} d x

Apply Trigonometric Substitution

: \int \frac{1}{4 sec ( u )} d u

= \int \frac{1}{4 sec ( 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{1}{sec ( u )} d u

Rewrite using trig identities

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

Use the common integral:

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

= \frac{1}{4} sin (u)

Substitute back

u = arcsec (\frac{1}{2} x)

= \frac{1}{4} sin (arcsec (\frac{1}{2} x))

Simplify

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

= \frac{x ^{2} - 4}{4 x}

Add a constant to the solution

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

x \to 0 + lim (x^{12} ln (x))

\frac{d}{d x} (ln (x^{2} + 3 x + 3))

f (x) = 4 x^{4} + 3 x^{2} - 1

\frac{d y}{d x} = 4 y x

\int cos^{7} (x) d x

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