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

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

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

integral

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{3 ( 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}{3} \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}{3} arctan (u)

Substitute back

u = \frac{x}{3}

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

Add a constant to the solution

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

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

- sin^{2} (x) + tan^{2} (x)

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

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

\frac{( e ^{x} ) ^{2}}{2}

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