What is the integral of 6/(9x^2+1) ?

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

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integral of 6/(9x^2+1)

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

2 arctan (3 x) + C

Solution steps

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

Apply Integral Substitution

= 6 \cdot \int \frac{1}{3 ( 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}{3} \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}{3} arctan (u)

Substitute back

u = 3 x

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

Simplify

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

= 2 arctan (3 x)

Add a constant to the solution

= 2 arctan (3 x) + C

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