What is the integral of 1/(4+81x^2) ?

The integral of 1/(4+81x^2) is 1/18 arctan(9/2 x)+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+81x^2)

\int \frac{1}{4 + 81 x ^{2}} d x

\frac{1}{18} arctan (\frac{9}{2} x) + C

Solution steps

\int \frac{1}{4 + 81 x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{18 ( 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}{18} \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}{18} arctan (u)

Substitute back

u = \frac{9}{2} x

= \frac{1}{18} arctan (\frac{9}{2} x)

Add a constant to the solution

= \frac{1}{18} arctan (\frac{9}{2} x) + C

\int (3 + 2 x y) d x

\int e^{x} - cos (x) + 6 x + 4 d x

\int - 16 sin (- 4 t) d t

\int \frac{3 x + 2}{2 x} d x

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

What is the integral of 1/(4+81x^2) ?
The integral of 1/(4+81x^2) is 1/18 arctan(9/2 x)+C