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

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

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integral of 1/(x^2+2)

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

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

Solution steps

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

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{x}{2}

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{x - 2}{y - 4}

\int x^{3} + 7 x - 5 x e^{- 2 x^{2}} d x

\int \frac{1}{x ^{2} + 18 x} d x

\frac{d}{d t} (tan (t))

x \to 3 lim (x - 2)

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