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

\int \frac{- 3 x ^{3} + 2 x ^{2} - 28 x + 10}{x ^{2} + 9} d x

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

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

f (x) = lo g_{e} (x)

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

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