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

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

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integral of v/(1+v^2)

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

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

Solution steps

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

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 = 1 + v^{2}

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

Add a constant to the solution

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

\int sinh (x) cosh (x) d x

1 - x, 0

y^{^{'''}} + 3 y^{^{''}} - 4 y = 0

y^{^{''}} + 3 y^{^{'}} + 2 y = cos (2 x)

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