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

The integral of (((sqrt(s)-2)^2)/(4s^2)) is 1/4 ln|s|+2/(sqrt(s))-1/s+C

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

\int (\frac{( s - 2 ) ^{2}}{4 s ^{2}}) d s

\frac{1}{4} ln ∣ s ∣ + \frac{2}{s} - \frac{1}{s} + C

Solution steps

\int \frac{( s - 2 ) ^{2}}{4 s ^{2}} d s

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{( s - 2 ) ^{2}}{s ^{2}} d s

Expand

\frac{( s - 2 ) ^{2}}{s ^{2}} : \frac{1}{s} - \frac{4}{s ^{\frac{3}{2}}} + \frac{4}{s ^{2}}

= \frac{1}{4} \cdot \int \frac{1}{s} - \frac{4}{s ^{\frac{3}{2}}} + \frac{4}{s ^{2}} 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

= \frac{1}{4} (\int \frac{1}{s} d s - \int \frac{4}{s ^{\frac{3}{2}}} d s + \int \frac{4}{s ^{2}} d s)

\int \frac{1}{s} d s = ln ∣ s ∣

\int \frac{4}{s ^{\frac{3}{2}}} d s = - \frac{8}{s}

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

= \frac{1}{4} (ln ∣ s ∣ - (- \frac{8}{s}) - \frac{4}{s})

Simplify

\frac{1}{4} (ln ∣ s ∣ - (- \frac{8}{s}) - \frac{4}{s}) : \frac{1}{4} ln ∣ s ∣ + \frac{2}{s} - \frac{1}{s}

= \frac{1}{4} ln ∣ s ∣ + \frac{2}{s} - \frac{1}{s}

Add a constant to the solution

= \frac{1}{4} ln ∣ s ∣ + \frac{2}{s} - \frac{1}{s} + C

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What is the integral of (((sqrt(s)-2)^2)/(4s^2)) ?
The integral of (((sqrt(s)-2)^2)/(4s^2)) is 1/4 ln|s|+2/(sqrt(s))-1/s+C