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

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

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

integral

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

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

Solution steps

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

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

Add a constant to the solution

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

4 x^{3} + 3 x^{2} + 2

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

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

0.5 ((1 - cos (2 t)))

\frac{e ^{\frac{1}{x}}}{x}

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