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

The integral of (3x)/(6x^2+9) is 1/4 ln|6x^2+9|+C

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integral of (3x)/(6x^2+9)

\int \frac{3 x}{6 x ^{2} + 9} d x

\frac{1}{4} ln 6 x^{2} + 9 + C

Solution steps

\int \frac{3 x}{6 x ^{2} + 9} 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}{6 x ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{12 u} d u

= 3 \cdot \int \frac{1}{12 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}{12} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{12} ln ∣ u ∣

Substitute back

u = 6 x^{2} + 9

= 3 \cdot \frac{1}{12} ln 6 x^{2} + 9

Simplify

3 \cdot \frac{1}{12} ln 6 x^{2} + 9 : \frac{1}{4} ln 6 x^{2} + 9

= \frac{1}{4} ln 6 x^{2} + 9

Add a constant to the solution

= \frac{1}{4} ln 6 x^{2} + 9 + C

f (x) = - \frac{2 x - 10}{x - 3}, at x = 2

y = x^{2}, x = y^{2}, y = 1

\frac{d}{d x} (\frac{x ^{3}}{x ^{3}})

\int \frac{1}{w} d w

y = x^{2} - 9, y = - x + 3, x = 0, x = 2

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