What is the integral of 1/(x^2+16) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(x^2+16)

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

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

Solution steps

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

Apply Integral Substitution

= \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

= \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)

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

Substitute back

u = \frac{x}{4}

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

Add a constant to the solution

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

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