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

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

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integral of 1/(sqrt(v^2+1))

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

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

Solution steps

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

Use the common integral:

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

= ln v^{2} + 1 + v

Add a constant to the solution

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

y = 1 - x - x^{2}

\int \frac{x ^{4}}{4} + cosh (x) + C d x

y = e^{2} + 2 x

x \to 0 lim (\frac{e ^{x} - 1}{15 x})

\frac{d}{d x} (\frac{e ^{a x} - e ^{- a x}}{e ^{a x} + e ^{- a x}})

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