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

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

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

integral

\frac{1}{( 2 x ^{2} + 3 )}

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

Solution steps

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

Apply Integral Substitution

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

Substitute back

u = \frac{2}{3} x

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

Combine same powers :

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

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

Add a constant to the solution

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

x + 3

\frac{4 x ^{2}}{x ^{2} + 2}

a^{2} - x^{2}

\frac{e ^{x}}{( 1 + e ^{x} )}

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

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