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

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

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

integral

\frac{s}{( s ^{2} + 1 )}

\frac{1}{2} ln s^{2} + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = s^{2} + 1

= \frac{1}{2} ln s^{2} + 1

Add a constant to the solution

= \frac{1}{2} ln s^{2} + 1 + C

x^{1}

\frac{x ^{4}}{( x ^{3} + 1 )}

\frac{x}{3}

sin (x^{4})

\frac{1}{( 1 + y ^{2} )}

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