What is the integral of (6(1+s))/(s(s^2+7s+6)) ?

The integral of (6(1+s))/(s(s^2+7s+6)) is ln|s|-ln|s+6|+C

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integral of (6(1+s))/(s(s^2+7s+6))

\int \frac{6 ( 1 + s )}{s ( s ^{2} + 7 s + 6 )} d s

ln ∣ s ∣ - ln ∣ s + 6 ∣ + C

Solution steps

\int \frac{6 ( 1 + s )}{s ( s ^{2} + 7 s + 6 )} d s

Take the constant out:

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

= 6 \cdot \int \frac{1 + s}{s ( s ^{2} + 7 s + 6 )} d s

Take the partial fraction of

\frac{1 + s}{s ( s ^{2} + 7 s + 6 )} : \frac{1}{6 s} - \frac{1}{6 ( s + 6 )}

= 6 \cdot \int \frac{1}{6 s} - \frac{1}{6 ( s + 6 )} 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

= 6 (\int \frac{1}{6 s} d s - \int \frac{1}{6 ( s + 6 )} d s)

\int \frac{1}{6 s} d s = \frac{1}{6} ln ∣ s ∣

\int \frac{1}{6 ( s + 6 )} d s = \frac{1}{6} ln ∣ s + 6 ∣

= 6 (\frac{1}{6} ln ∣ s ∣ - \frac{1}{6} ln ∣ s + 6 ∣)

Simplify

6 (\frac{1}{6} ln ∣ s ∣ - \frac{1}{6} ln ∣ s + 6 ∣) : ln ∣ s ∣ - ln ∣ s + 6 ∣

= ln ∣ s ∣ - ln ∣ s + 6 ∣

Add a constant to the solution

= ln ∣ s ∣ - ln ∣ s + 6 ∣ + C

\int 1 - z^{2}

\int \frac{x ^{3} + x}{x - 1} d x

h \to 0 lim (\frac{1}{h})

y = (3 - x^{2}) (x^{3} - 3 x + 2)

\int_{5}^{8} \frac{1}{x ^{2} - 1} d x

What is the integral of (6(1+s))/(s(s^2+7s+6)) ?
The integral of (6(1+s))/(s(s^2+7s+6)) is ln|s|-ln|s+6|+C