What is the integral of (sqrt(x^2-4))/(3x^3) ?

The integral of (sqrt(x^2-4))/(3x^3) is 1/3 (-(sqrt(x^2-4))/(2x^2)+1/4 arcsec(1/2 x))+C

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integral of (sqrt(x^2-4))/(3x^3)

\int \frac{x ^{2} - 4}{3 x ^{3}} d x

\frac{1}{3} (- \frac{x ^{2} - 4}{2 x ^{2}} + \frac{1}{4} arcsec (\frac{1}{2} x)) + C

Solution steps

\int \frac{x ^{2} - 4}{3 x ^{3}} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{x ^{2} - 4}{x ^{3}} d x

Apply Integration By Parts

= \frac{1}{3} (- \frac{x ^{2} - 4}{2 x ^{2}} - \int - \frac{1}{2 x x ^{2} - 4} d x)

\int - \frac{1}{2 x x ^{2} - 4} d x = - \frac{1}{4} arcsec (\frac{1}{2} x)

= \frac{1}{3} (- \frac{x ^{2} - 4}{2 x ^{2}} - (- \frac{1}{4} arcsec (\frac{1}{2} x)))

Simplify

= \frac{1}{3} (- \frac{x ^{2} - 4}{2 x ^{2}} + \frac{1}{4} arcsec (\frac{1}{2} x))

Add a constant to the solution

= \frac{1}{3} (- \frac{x ^{2} - 4}{2 x ^{2}} + \frac{1}{4} arcsec (\frac{1}{2} x)) + C

What is the integral of (sqrt(x^2-4))/(3x^3) ?
The integral of (sqrt(x^2-4))/(3x^3) is 1/3 (-(sqrt(x^2-4))/(2x^2)+1/4 arcsec(1/2 x))+C