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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(4+x^2)

integral

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

\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

ln (x^{3})

((9^{9})^{9})^{x}

(sin^{2} (x))^{2}

x^{2} \cdot e^{- x}

\frac{1}{( 1 + sin ^{2} ( x ) )}

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