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

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

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

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

Solution steps

\int \frac{1}{2 + 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 = \frac{x}{2}

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

Add a constant to the solution

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

\int 10 - x^{2} (5 x) d x

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

\int 3 (3 x - 5)^{4} d x

\int \frac{x ^{3}}{x + 1} d x

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

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