What is the integral of 1/(4+9x^2) ?

The integral of 1/(4+9x^2) is 1/6 arctan(3/2 x)+C

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integral of 1/(4+9x^2)

\int \frac{1}{4 + 9 x ^{2}} d x

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

Solution steps

\int \frac{1}{4 + 9 x ^{2}} 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{3}{2} x

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

Add a constant to the solution

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

\frac{d y}{d x} - 4 y = x^{2} + 9

y = 8 x - x^{2} (1.7)

x \to 0 lim (e^{x} + \frac{1}{x})

\frac{d}{d x} (0.8 x^{1.3})

x \to 3 - lim (- x + 5)

What is the integral of 1/(4+9x^2) ?
The integral of 1/(4+9x^2) is 1/6 arctan(3/2 x)+C