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

The integral of ()/(sqrt(4s^2+8s+19)) is (ds)/(sqrt(19+4s^2+8s))x+C

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

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

\frac{d s}{19 + 4 s ^{2} + 8 s} x + C

Solution steps

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

Integral of a constant:

\int a d x = a x

= \frac{d s}{19 + 4 s ^{2} + 8 s} x

Add a constant to the solution

= \frac{d s}{19 + 4 s ^{2} + 8 s} x + C

\int \frac{x ^{2} - 6}{x ( x - 3 )} d x

\int \frac{8 x}{2 x ^{2} + 1} d x

\int \frac{7 x ^{2} - 9 x + 6}{( x - 1 ) ( x ^{2} + 1 )} d x

\int \frac{ln ( 4 x ^{2} )}{x} d x

\int 19 x sin (x) cos (x) d x

What is the integral of ()/(sqrt(4s^2+8s+19)) ?
The integral of ()/(sqrt(4s^2+8s+19)) is (ds)/(sqrt(19+4s^2+8s))x+C