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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

= 2 \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

= 2 \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)

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

Substitute back

u = \frac{2}{3} x

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{9 x - 1} d x

\int \frac{x - 2}{1 - x ^{2}} d x

\int (1 - x^{2} - 3 x^{5}) d x

\int e^{u + 1} d u

\int \frac{4 x}{1 - 4 x ^{4}} d x

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