What is the integral of 1/(3x^2+4) ?

The integral of 1/(3x^2+4) is 1/(2sqrt(3))arctan((sqrt(3))/2 x)+C

integral of 1/(3x^2+4)

\int \frac{1}{3 x ^{2} + 4} d x

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

Solution steps

\int \frac{1}{3 x ^{2} + 4} d x

Apply Integral Substitution

= \int \frac{1}{2 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}{2 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}{2 3} arctan (u)

Substitute back

u = \frac{3}{2} x

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

Add a constant to the solution

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

t an g e n t y = 3 x + 6 cos (x), (\frac{π}{3}, π + 3)

d er i v a t i v e y = (x^{2} + 8)^{2} - 4 x

\frac{d}{d x} (\frac{x ^{5}}{10})

x \to \infty lim (4 x + 2)

\int ln (x) + sin (π x) d x

What is the integral of 1/(3x^2+4) ?
The integral of 1/(3x^2+4) is 1/(2sqrt(3))arctan((sqrt(3))/2 x)+C