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

y = 2 x^{3}, (- 3, - 54)

x \to - 5 lim (\frac{x + 4}{x + 5})

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

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

\frac{\partial}{\partial x} (ln (2 x + 3 y))

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