What is the integral of 6/(9x^2+4) ?

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

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integral of 6/(9x^2+4)

\int \frac{6}{9 x ^{2} + 4} d x

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

Solution steps

\int \frac{6}{9 x ^{2} + 4} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{9 x ^{2} + 4} d x

Apply Integral Substitution

= 6 \cdot \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

= 6 \cdot \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)

= 6 \cdot \frac{1}{6} arctan (u)

Substitute back

u = \frac{3}{2} x

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

Simplify

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

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

Add a constant to the solution

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

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