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

x \to 0 lim (\frac{3 x}{2 x ^{2} - x})

\frac{( x + 4 ) ( x ^{2} - 4 x + 16 )}{x ^{3}}

\int \frac{1}{3} e^{2 - x} d x

\frac{d}{d x} (x 36 - x^{2})

\frac{d}{d x} ((2 x + 3) (5 x^{2} - 4 x))

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