What is the integral of 1/(sqrt(4s^2+8s+19)) ?

The integral of 1/(sqrt(4s^2+8s+19)) is 1/2 ln| 2/(sqrt(15))(s+1)+sqrt((4s^2+8s+19)/(15))|+C

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integral of 1/(sqrt(4s^2+8s+19))

\int \frac{1}{4 s ^{2} + 8 s + 19} d s

\frac{1}{2} ln \frac{2}{15} (s + 1) + \frac{4 s ^{2} + 8 s + 19}{15} + C

Solution steps

\int \frac{1}{4 s ^{2} + 8 s + 19} d s

Complete the square

4 s^{2} + 8 s + 19 : 4 (s + 1)^{2} + 15

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{2} d v

= \int \frac{sec ( v )}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{2} ln tan (arctan (\frac{2}{15} (s + 1))) + sec (arctan (\frac{2}{15} (s + 1)))

Simplify

\frac{1}{2} ln tan (arctan (\frac{2}{15} (s + 1))) + sec (arctan (\frac{2}{15} (s + 1))) : \frac{1}{2} ln \frac{2}{15} (s + 1) + \frac{4 s ^{2} + 8 s + 19}{15}

= \frac{1}{2} ln \frac{2}{15} (s + 1) + \frac{4 s ^{2} + 8 s + 19}{15}

Add a constant to the solution

= \frac{1}{2} ln \frac{2}{15} (s + 1) + \frac{4 s ^{2} + 8 s + 19}{15} + C

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What is the integral of 1/(sqrt(4s^2+8s+19)) ?
The integral of 1/(sqrt(4s^2+8s+19)) is 1/2 ln| 2/(sqrt(15))(s+1)+sqrt((4s^2+8s+19)/(15))|+C