What is the integral of 2/(16+x^2) ?

The integral of 2/(16+x^2) is 1/2 arctan(x/4)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2/(16+x^2)

\int \frac{2}{16 + x ^{2}} d x

\frac{1}{2} arctan (\frac{x}{4}) + C

Solution steps

\int \frac{2}{16 + x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{16 + x ^{2}} d x

Apply Integral Substitution

= 2 \cdot \int \frac{1}{4 ( u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 2 \cdot \frac{1}{4} arctan (u)

Substitute back

u = \frac{x}{4}

= 2 \cdot \frac{1}{4} arctan (\frac{x}{4})

Simplify

2 \cdot \frac{1}{4} arctan (\frac{x}{4}) : \frac{1}{2} arctan (\frac{x}{4})

= \frac{1}{2} arctan (\frac{x}{4})

Add a constant to the solution

= \frac{1}{2} arctan (\frac{x}{4}) + C

\int (\frac{1}{x 9 x ^{2} - 1}) d x

\int 6 x sin (2 x) d x

\int \frac{( e ^{2 x} - 2 e ^{- x} + 1 )}{e ^{x}} d x

\int x^{3} cos (- \frac{3}{4} x) d x

\int \frac{x ^{2}}{( 2 x + 3 ) ( 4 x ^{2} - 1 )} d x

What is the integral of 2/(16+x^2) ?
The integral of 2/(16+x^2) is 1/2 arctan(x/4)+C