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

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

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integral of 1/(1+v^2)

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

arctan (v) + C

Solution steps

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

Use the common integral:

\int \frac{1}{1 + v ^{2}} d v = arctan (v)

= arctan (v)

Add a constant to the solution

= arctan (v) + C

x \to 2 lim (- 4)

\int x e^{- x (1 + y)} d x

l a pl a ce t r an s f or m e^{2 t} t sin (5 t)

n = 0 \sum \infty (- 1)^{n} x^{2 n}

t an g e n t y = \frac{3 x + 1}{x ^{2} + 1}, (1, 2)