What is the integral of (10s+10)/((s^2+1)(s-1)^3) ?

The integral of (10s+10)/((s^2+1)(s-1)^3) is 5arctan(s)+5/(s-1)-5/((s-1)^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (10s+10)/((s^2+1)(s-1)^3)

\int \frac{10 s + 10}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

5 arctan (s) + \frac{5}{s - 1} - \frac{5}{( s - 1 ) ^{2}} + C

Solution steps

\int \frac{10 s + 10}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

Take the partial fraction of

\frac{10 s + 10}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} : \frac{5}{s ^{2} + 1} - \frac{5}{( s - 1 ) ^{2}} + \frac{10}{( s - 1 ) ^{3}}

= \int \frac{5}{s ^{2} + 1} - \frac{5}{( s - 1 ) ^{2}} + \frac{10}{( s - 1 ) ^{3}} 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{5}{s ^{2} + 1} d s - \int \frac{5}{( s - 1 ) ^{2}} d s + \int \frac{10}{( s - 1 ) ^{3}} d s

\int \frac{5}{s ^{2} + 1} d s = 5 arctan (s)

\int \frac{5}{( s - 1 ) ^{2}} d s = - \frac{5}{s - 1}

\int \frac{10}{( s - 1 ) ^{3}} d s = - \frac{5}{( s - 1 ) ^{2}}

= 5 arctan (s) - (- \frac{5}{s - 1}) - \frac{5}{( s - 1 ) ^{2}}

Simplify

= 5 arctan (s) + \frac{5}{s - 1} - \frac{5}{( s - 1 ) ^{2}}

Add a constant to the solution

= 5 arctan (s) + \frac{5}{s - 1} - \frac{5}{( s - 1 ) ^{2}} + C

x - 1

\int sinh^{5} (x) d x

\int \frac{x ^{2} - x ^{5} + x}{x ^{5}} d x

\int \frac{3 e ^{3 x}}{x + 10} d x

\int cos^{5} (3 x) sin^{3} (3 x) d x

What is the integral of (10s+10)/((s^2+1)(s-1)^3) ?
The integral of (10s+10)/((s^2+1)(s-1)^3) is 5arctan(s)+5/(s-1)-5/((s-1)^2)+C