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

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

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

integral

\frac{1}{( 9 x ^{2} + 4 )}

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

Solution steps

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

y = \frac{1}{x ^{\frac{1}{2}}}

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

\frac{e ^{2} x - 1}{e ^{2} x + 1}

3 x + 1

x (cos (x))^{2}

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