What is the integral of 1/(s+sqrt(2)) ?

The integral of 1/(s+sqrt(2)) is ln|s+sqrt(2)|+C

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integral of 1/(s+sqrt(2))

\int \frac{1}{s + 2} d s

ln s + 2 + C

Solution steps

\int \frac{1}{s + 2} d s

Apply u-substitution

: \int \frac{1}{u} d u

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = s + 2

= ln s + 2

Add a constant to the solution

= ln s + 2 + C

(- 2, - 5), (2, - 8)

y = (1 + 5 x)^{11}, (0, 1)

\int_{4}^{8} \frac{1}{x ^{2} - 1} d x

- \frac{2}{( 1 - x ) ^{3}}

\int \frac{x ^{3}}{( x ^{2} + 5 ) ^{2}} d x

What is the integral of 1/(s+sqrt(2)) ?
The integral of 1/(s+sqrt(2)) is ln|s+sqrt(2)|+C