What is the integral of (s^2-4s-6)/(s^4) ?

The integral of (s^2-4s-6)/(s^4) is -1/s+2/(s^2)+2/(s^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (s^2-4s-6)/(s^4)

\int \frac{s ^{2} - 4 s - 6}{s ^{4}} d s

- \frac{1}{s} + \frac{2}{s ^{2}} + \frac{2}{s ^{3}} + C

Solution steps

\int \frac{s ^{2} - 4 s - 6}{s ^{4}} d s

Expand

\frac{s ^{2} - 4 s - 6}{s ^{4}} : \frac{1}{s ^{2}} - \frac{4}{s ^{3}} - \frac{6}{s ^{4}}

= \int \frac{1}{s ^{2}} - \frac{4}{s ^{3}} - \frac{6}{s ^{4}} d s

Apply the Sum Rule:

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

= \int \frac{1}{s ^{2}} d s - \int \frac{4}{s ^{3}} d s - \int \frac{6}{s ^{4}} d s

\int \frac{1}{s ^{2}} d s = - \frac{1}{s}

\int \frac{4}{s ^{3}} d s = - \frac{2}{s ^{2}}

\int \frac{6}{s ^{4}} d s = - \frac{2}{s ^{3}}

= - \frac{1}{s} - (- \frac{2}{s ^{2}}) - (- \frac{2}{s ^{3}})

Simplify

= - \frac{1}{s} + \frac{2}{s ^{2}} + \frac{2}{s ^{3}}

Add a constant to the solution

= - \frac{1}{s} + \frac{2}{s ^{2}} + \frac{2}{s ^{3}} + C

\int \frac{1}{2} x e^{\frac{x}{4}} d x

\int (7^{x} + 1 sin (10 x) + 3 sec^{2} (x)) d x

\int \frac{x + 1}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

\int x^{2} a^{2} + x^{2} d x

\int \frac{1}{1 - 16 y ^{2}} d y

What is the integral of (s^2-4s-6)/(s^4) ?
The integral of (s^2-4s-6)/(s^4) is -1/s+2/(s^2)+2/(s^3)+C