What is the integral of 1/(s^2+3s+6) ?

The integral of 1/(s^2+3s+6) is 2/(sqrt(15))arctan(2/(sqrt(15))s+sqrt(3/5))+C

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integral of 1/(s^2+3s+6)

\int \frac{1}{s ^{2} + 3 s + 6} d s

\frac{2}{15} arctan (\frac{2}{15} s + \frac{3}{5}) + C

Solution steps

\int \frac{1}{s ^{2} + 3 s + 6} d s

Complete the square

s^{2} + 3 s + 6 : (s + \frac{3}{2})^{2} + \frac{15}{4}

= \int \frac{1}{( s + \frac{3}{2} ) ^{2} + \frac{15}{4}} d s

Apply u-substitution

: \int \frac{4}{4 u ^{2} + 15} d u

= \int \frac{4}{4 u ^{2} + 15} d u

Take the constant out:

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

= 4 \cdot \int \frac{1}{4 u ^{2} + 15} d u

Apply Integral Substitution

= 4 \cdot \int \frac{1}{2 15 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2 15} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 4 \cdot \frac{1}{2 15} arctan (v)

Substitute back

= 4 \cdot \frac{1}{2 15} arctan (\frac{2}{15} (s + \frac{3}{2}))

Simplify

4 \cdot \frac{1}{2 15} arctan (\frac{2}{15} (s + \frac{3}{2})) : \frac{2}{15} arctan (\frac{2}{15} s + \frac{3}{5})

= \frac{2}{15} arctan (\frac{2}{15} s + \frac{3}{5})

Add a constant to the solution

= \frac{2}{15} arctan (\frac{2}{15} s + \frac{3}{5}) + C

\frac{d}{d x} (tan^{- 1} (x))

2 x, x^{2}, 0, 2

\int \frac{1}{2} e^{- 2 x} d x

(1, - 8), (5, 4)

x \to o lim (\frac{sin ( x )}{x})

What is the integral of 1/(s^2+3s+6) ?
The integral of 1/(s^2+3s+6) is 2/(sqrt(15))arctan(2/(sqrt(15))s+sqrt(3/5))+C