What is the integral of (6s^2-2)/(s^3) ?

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

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integral of (6s^2-2)/(s^3)

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

6 ln ∣ s ∣ + \frac{1}{s ^{2}} + C

Solution steps

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

Expand

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

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

\int \frac{6}{s} d s = 6 ln ∣ s ∣

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

= 6 ln ∣ s ∣ - (- \frac{1}{s ^{2}})

Simplify

= 6 ln ∣ s ∣ + \frac{1}{s ^{2}}

Add a constant to the solution

= 6 ln ∣ s ∣ + \frac{1}{s ^{2}} + C

\frac{1}{2 x + 5}

2 x^{2} - 3 x + 5

f (x) = \frac{sin ( 5 x )}{3 x}

\int 16 x^{5} d x

\int (8 x (x^{7} - \frac{1}{8})) d x

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