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

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

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integral of (3x)/(9x^2+1)

\int \frac{3 x}{9 x ^{2} + 1} d x

\frac{1}{6} ln 9 x^{2} + 1 + C

Solution steps

\int \frac{3 x}{9 x ^{2} + 1} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{9 x ^{2} + 1} d x

Apply u-substitution

= 3 \cdot \int \frac{1}{18 u} d u

Take the constant out:

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

= 3 \cdot \frac{1}{18} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{18} ln ∣ u ∣

Substitute back

u = 9 x^{2} + 1

= 3 \cdot \frac{1}{18} ln 9 x^{2} + 1

Simplify

3 \cdot \frac{1}{18} ln 9 x^{2} + 1 : \frac{1}{6} ln 9 x^{2} + 1

= \frac{1}{6} ln 9 x^{2} + 1

Add a constant to the solution

= \frac{1}{6} ln 9 x^{2} + 1 + C

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