What is the integral of (s^4+81)/(s(s^2+9)^2) ?

The integral of (s^4+81)/(s(s^2+9)^2) is 9/(s^2+9)+ln(s)+C

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integral of (s^4+81)/(s(s^2+9)^2)

\int \frac{s ^{4} + 81}{s ( s ^{2} + 9 ) ^{2}} d s

\frac{9}{s ^{2} + 9} + ln (s) + C

Solution steps

\int \frac{s ^{4} + 81}{s ( s ^{2} + 9 ) ^{2}} d s

Expand

\frac{s ^{4} + 81}{s ( s ^{2} + 9 ) ^{2}} : \frac{s ^{4}}{s ( s ^{2} + 9 ) ^{2}} + \frac{81}{s ( s ^{2} + 9 ) ^{2}}

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{s ^{4}}{s ( s ^{2} + 9 ) ^{2}} d s + \int \frac{81}{s ( s ^{2} + 9 ) ^{2}} d s

\int \frac{s ^{4}}{s ( s ^{2} + 9 ) ^{2}} d s = \frac{1}{2} (ln s^{2} + 9 + \frac{9}{s ^{2} + 9})

\int \frac{81}{s ( s ^{2} + 9 ) ^{2}} d s = ln (s) - \frac{1}{2} ln s^{2} + 9 + \frac{9}{2 ( s ^{2} + 9 )}

= \frac{1}{2} (ln s^{2} + 9 + \frac{9}{s ^{2} + 9}) + ln (s) - \frac{1}{2} ln s^{2} + 9 + \frac{9}{2 ( s ^{2} + 9 )}

Simplify

\frac{1}{2} (ln s^{2} + 9 + \frac{9}{s ^{2} + 9}) + ln (s) - \frac{1}{2} ln s^{2} + 9 + \frac{9}{2 ( s ^{2} + 9 )} : \frac{9}{s ^{2} + 9} + ln (s)

= \frac{9}{s ^{2} + 9} + ln (s)

Add a constant to the solution

= \frac{9}{s ^{2} + 9} + ln (s) + C

What is the integral of (s^4+81)/(s(s^2+9)^2) ?
The integral of (s^4+81)/(s(s^2+9)^2) is 9/(s^2+9)+ln(s)+C