What is the integral of 1/(x^2sqrt(x^2-3)) ?

The integral of 1/(x^2sqrt(x^2-3)) is (sqrt(x^2-3))/(3x)+C

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integral of 1/(x^2sqrt(x^2-3))

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

\frac{x ^{2} - 3}{3 x} + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{3 sec ( u )} d u

= \int \frac{1}{3 sec ( u )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{3} sin (u)

Substitute back

u = arcsec (\frac{1}{3} x)

= \frac{1}{3} sin (arcsec (\frac{1}{3} x))

Simplify

\frac{1}{3} sin (arcsec (\frac{1}{3} x)) : \frac{x ^{2} - 3}{3 x}

= \frac{x ^{2} - 3}{3 x}

Add a constant to the solution

= \frac{x ^{2} - 3}{3 x} + C

x \to 0 lim (- x ln (x) - x)

\int x \cdot x d x

\frac{1}{x - 1}

x \cdot y^{^{'}} = x^{2} - 2 x y - y^{2} + y

\int 3 x 2 x^{2} + 1 d x

What is the integral of 1/(x^2sqrt(x^2-3)) ?
The integral of 1/(x^2sqrt(x^2-3)) is (sqrt(x^2-3))/(3x)+C