What is the integral of (sqrt(x^2-9))/(2x) ?

The integral of (sqrt(x^2-9))/(2x) is -3/2 arcsec(1/3 x)+1/2 sqrt(x^2-9)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sqrt(x^2-9))/(2x)

\int \frac{x ^{2} - 9}{2 x} d x

- \frac{3}{2} arcsec (\frac{1}{3} x) + \frac{1}{2} x^{2} - 9 + C

Solution steps

\int \frac{x ^{2} - 9}{2 x} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{x ^{2} - 9}{x} d x

Apply Trigonometric Substitution

: \int 3 tan^{2} (u) d u

= \frac{1}{2} \cdot \int 3 tan^{2} (u) d u

Take the constant out:

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

= \frac{1}{2} \cdot 3 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot 3 \cdot \int - 1 + sec^{2} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} \cdot 3 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

\int sec^{2} (u) d u = tan (u)

= \frac{1}{2} \cdot 3 (- u + tan (u))

Substitute back

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

= \frac{1}{2} \cdot 3 (- arcsec (\frac{1}{3} x) + tan (arcsec (\frac{1}{3} x)))

Simplify

\frac{1}{2} \cdot 3 (- arcsec (\frac{1}{3} x) + tan (arcsec (\frac{1}{3} x))) : - \frac{3}{2} arcsec (\frac{1}{3} x) + \frac{1}{2} x^{2} - 9

= - \frac{3}{2} arcsec (\frac{1}{3} x) + \frac{1}{2} x^{2} - 9

Add a constant to the solution

= - \frac{3}{2} arcsec (\frac{1}{3} x) + \frac{1}{2} x^{2} - 9 + C

\int 7 x 7 x^{2} + 14 d x

\int \frac{2 x ^{4}}{1 - 8 x ^{5}} d x

\int \frac{d}{1} y d x

\int - 3 sin^{2} (x) cos^{2} (x) d x

\int \frac{4 x ^{2}}{5 x ^{3} + 8} d x

What is the integral of (sqrt(x^2-9))/(2x) ?
The integral of (sqrt(x^2-9))/(2x) is -3/2 arcsec(1/3 x)+1/2 sqrt(x^2-9)+C