What is the integral of 1/(2+3x^2) ?

The integral of 1/(2+3x^2) is 1/(sqrt(6))arctan(sqrt(3/2)x)+C

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integral of 1/(2+3x^2)

\int \frac{1}{2 + 3 x ^{2}} d x

\frac{1}{6} arctan (\frac{3}{2} x) + C

Solution steps

\int \frac{1}{2 + 3 x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{6 ( 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}{6} \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}{6} arctan (u)

Substitute back

u = \frac{3}{2} x

= \frac{1}{6} arctan (\frac{3}{2} x)

Combine same powers :

\frac{x}{y} = \frac{x}{y}

= \frac{1}{6} arctan (\frac{3}{2} x)

Add a constant to the solution

= \frac{1}{6} arctan (\frac{3}{2} x) + C

\int ln (\frac{( 1 + x )}{1 - x}) d x

\frac{d ^{2}}{d x ^{2}} (2 sec (x))

f (x) = \frac{4 x}{x ^{2} + 1}

\frac{d}{d x} ((3 x - 5)^{2})

x \to 0 lim (\frac{5}{x ^{2} + x} - \frac{5}{x})

What is the integral of 1/(2+3x^2) ?
The integral of 1/(2+3x^2) is 1/(sqrt(6))arctan(sqrt(3/2)x)+C