What is the integral of (sqrt(w^2-25))/(w^3) ?

The integral of (sqrt(w^2-25))/(w^3) is -(sqrt(w^2-25))/(2w^2)+1/10 arcsec(1/5 w)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sqrt(w^2-25))/(w^3)

\int \frac{w ^{2} - 25}{w ^{3}} d w

- \frac{w ^{2} - 25}{2 w ^{2}} + \frac{1}{10} arcsec (\frac{1}{5} w) + C

Solution steps

\int \frac{w ^{2} - 25}{w ^{3}} d w

Apply Integration By Parts

: - \frac{w ^{2} - 25}{2 w ^{2}} - \int - \frac{1}{2 w w ^{2} - 25} d w

= - \frac{w ^{2} - 25}{2 w ^{2}} - \int - \frac{1}{2 w w ^{2} - 25} d w

\int - \frac{1}{2 w w ^{2} - 25} d w = - \frac{1}{10} arcsec (\frac{1}{5} w)

= - \frac{w ^{2} - 25}{2 w ^{2}} - (- \frac{1}{10} arcsec (\frac{1}{5} w))

Simplify

= - \frac{w ^{2} - 25}{2 w ^{2}} + \frac{1}{10} arcsec (\frac{1}{5} w)

Add a constant to the solution

= - \frac{w ^{2} - 25}{2 w ^{2}} + \frac{1}{10} arcsec (\frac{1}{5} w) + C

0.03 \cdot (1 - cos (2 t))

x \to 0 lim ((2 x^{2})^{x^{4}})

\frac{d}{d n} ((lo g_{10} (n))^{3})

\int \frac{1}{y ( 2 - y )} d y

f (x) = ln (\frac{1 - x}{1 + x})

What is the integral of (sqrt(w^2-25))/(w^3) ?
The integral of (sqrt(w^2-25))/(w^3) is -(sqrt(w^2-25))/(2w^2)+1/10 arcsec(1/5 w)+C