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

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

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integral of 1/(x^2+4)

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

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

Solution steps

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

\frac{d}{d x} (sin (8 x) cos (9 y))

(x^{x} + 1)^{2}

\int x (18 x + 4) d x

\frac{x ^{2}}{2 ( 1 - x )}

x \to - 2 lim (\frac{3}{x - 2})

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