What is the integral of 1/(4+x^2) ?

The integral of 1/(4+x^2) is 1/2 arctan(x/2)+C

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integral of 1/(4+x^2)

\int \frac{1}{4 + x ^{2}} d x

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

Solution steps

\int \frac{1}{4 + x ^{2}} d x

Apply Integral Substitution

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

Use the common integral:

\int \frac{1}{u ^{2} + 1} 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

f (x) = 2 x ln (3 x) - 2 x

\int (x - 1)^{2} x^{3} (2 - x) d x

x \to - \infty lim (\frac{4}{x} - 12)

\frac{d}{d x} (2 sec^{2} (x) \cdot tan (x))

x d x + (x^{2} y + 4 y) d y = 0

What is the integral of 1/(4+x^2) ?
The integral of 1/(4+x^2) is 1/2 arctan(x/2)+C