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

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

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integral of 1/(y^2+2)

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{2 ( 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}{2} \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}{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

\frac{\partial}{\partial x} (12 - x^{2} - y^{2})

y^{^{'}} = k y (100 - y)

\frac{d}{d x} (11 sin (x))

\frac{\partial}{\partial x} (2 y + z^{3} - x z)

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

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