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

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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

f (x) = - 2 x^{4} + 10 x^{2}, at x = - 1

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

\int x^{100} + x^{99} d x

(x^{2} - y^{2}) d x + 2 x y d y = 0

\int_{0}^{1} t^{3} (3 + t^{4})^{3} d t

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