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

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

Popular >

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

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

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

Solution steps

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

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

Add a constant to the solution

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

\int \frac{2}{5} - \frac{9}{x} d x

x d y - (y + x^{3} e^{x^{2}}) d x = 0

\int cot (10 x) csc^{4} (10 x) d x

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 5)^{7})

\frac{\partial}{\partial x} (2 x + y)

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