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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integral Substitution

= \frac{1}{9} \cdot \int \frac{3}{2 ( 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}{9} \cdot \frac{3}{2} \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}{9} \cdot \frac{3}{2} arctan (u)

Substitute back

u = \frac{x}{\frac{2}{3}}

= \frac{1}{9} \cdot \frac{3}{2} arctan (\frac{x}{\frac{2}{3}})

Simplify

\frac{1}{9} \cdot \frac{3}{2} arctan (\frac{x}{\frac{2}{3}}) : \frac{1}{6} arctan (\frac{3 x}{2})

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

Add a constant to the solution

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

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