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

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

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

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

\frac{1}{2} ln ∣ 2 x + 3 ∣ + C

Solution steps

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

Simplify

\frac{x}{6 x ^{2} + 9 x} : \frac{1}{3 ( 2 x + 3 )}

= 3 \cdot \int \frac{1}{3 ( 2 x + 3 )} d x

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= 3 \cdot \frac{1}{3} \cdot \int \frac{1}{2 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}{3} \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{3} \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x + 3

= 3 \cdot \frac{1}{3} \cdot \frac{1}{2} ln ∣ 2 x + 3 ∣

Simplify

3 \cdot \frac{1}{3} \cdot \frac{1}{2} ln ∣ 2 x + 3 ∣ : \frac{1}{2} ln ∣ 2 x + 3 ∣

= \frac{1}{2} ln ∣ 2 x + 3 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 x + 3 ∣ + C

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