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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((1+4x^2))

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

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

Solution steps

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

Apply Integral Substitution

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

Use the common integral:

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

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

Substitute back

u = 2 x

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

Add a constant to the solution

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

\frac{d x}{d t} - 3 x = cos (4 t)

x \to 2 - lim (\frac{4}{( x - 2 ) ^{3}})

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

\frac{d}{d x} ((sinh (x^{2}))^{\frac{1}{x}})

\frac{\partial}{\partial y} (5 x^{4} y)

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