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

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

\int \frac{1}{2 + y ^{2}} d y

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

Solution steps

\int \frac{1}{2 + y ^{2}} d y

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{y}{2}

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

Add a constant to the solution

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

cot^{6} (x)

y^{^{''}} + 25 y = t sin (2 t)

\frac{d}{d x} (0.5 e^{- 0.5 (x - 1)})

81 y^{^{''}} + 144 y^{^{'}} + 64 y = 0

\frac{\partial}{\partial x} (\frac{1}{2 cos ^{2} ( x )})

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