What is the integral of (sqrt(4x^2-9))/(x^3) ?

The integral of (sqrt(4x^2-9))/(x^3) is -(sqrt(4x^2-9))/(2x^2)+2/3 arcsec(2/3 x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sqrt(4x^2-9))/(x^3)

\int \frac{4 x ^{2} - 9}{x ^{3}} d x

- \frac{4 x ^{2} - 9}{2 x ^{2}} + \frac{2}{3} arcsec (\frac{2}{3} x) + C

Solution steps

\int \frac{4 x ^{2} - 9}{x ^{3}} d x

Apply Integration By Parts

: - \frac{4 x ^{2} - 9}{2 x ^{2}} - \int - \frac{2}{x 4 x ^{2} - 9} d x

= - \frac{4 x ^{2} - 9}{2 x ^{2}} - \int - \frac{2}{x 4 x ^{2} - 9} d x

\int - \frac{2}{x 4 x ^{2} - 9} d x = - \frac{2}{3} arcsec (\frac{2}{3} x)

= - \frac{4 x ^{2} - 9}{2 x ^{2}} - (- \frac{2}{3} arcsec (\frac{2}{3} x))

Simplify

= - \frac{4 x ^{2} - 9}{2 x ^{2}} + \frac{2}{3} arcsec (\frac{2}{3} x)

Add a constant to the solution

= - \frac{4 x ^{2} - 9}{2 x ^{2}} + \frac{2}{3} arcsec (\frac{2}{3} x) + C

x \to 0 lim (x)

x \to 1 lim (\frac{1}{∣ x - 1 ∣})

\frac{\partial}{\partial y} (tan (3 + 2 x^{2} y^{3} z^{2}))

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

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

What is the integral of (sqrt(4x^2-9))/(x^3) ?
The integral of (sqrt(4x^2-9))/(x^3) is -(sqrt(4x^2-9))/(2x^2)+2/3 arcsec(2/3 x)+C