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

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

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integral of 1/(2+x^2)

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

\frac{1}{2} arctan (\frac{x}{2}) + C

Solution steps

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

Apply Integral Substitution

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

Use the common integral:

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

= \frac{1}{2} arctan (u)

Substitute back

u = \frac{x}{2}

= \frac{1}{2} arctan (\frac{x}{2})

Add a constant to the solution

= \frac{1}{2} arctan (\frac{x}{2}) + C

x \to 4 lim (\frac{x ^{2} - 16}{2 x ^{2} + 7 x - 4})

sin (2 x) d x + cos (5 y) d y = 0

\frac{d}{d x} (ln (x^{2}) + y^{2} - 1)

\frac{d}{d x} ((\frac{( 3 x + 1 )}{x ^{2}})^{3})

x \to + 0 lim (x (x - [x]))

What is the integral of 1/(2+x^2) ?
The integral of 1/(2+x^2) is 1/(sqrt(2))arctan(x/(sqrt(2)))+C