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)

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

\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

\int_{- x}^{x} t^{3} d t

\frac{d}{d x} (2 x^{3} - 4 x + 5)

d er i v a t i v e \frac{t}{t - 1}

(1 + cos (t)) y^{^{'}} = (1 + e - y) sin (t)

y^{^{''}} + 4 y = 8 tan (2 x)

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