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

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

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

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

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

Solution steps

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

Use the common integral:

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

= ln v^{2} + 1 + v

Add a constant to the solution

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

\int \frac{( ln ( x ) ) ^{5}}{x} d x

\frac{d}{d x} (5 sec (5 x) tan (5 x))

\int_{0}^{0.5} 8 y^{2} d y

\frac{5}{t ^{2}}

\int \frac{3}{9 x ^{2} + 6 x + 1} d x

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