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

The integral of (9x)/(3x+1) is 3x+1-ln|3x+1|+C

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

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

3 x + 1 - ln ∣ 3 x + 1 ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u - 1}{9 u} d u

= 9 \cdot \int \frac{u - 1}{9 u} d u

Take the constant out:

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

= 9 \cdot \frac{1}{9} \cdot \int \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= 9 \cdot \frac{1}{9} \cdot \int 1 - \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 9 \cdot \frac{1}{9} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 9 \cdot \frac{1}{9} (u - ln ∣ u ∣)

Substitute back

u = 3 x + 1

= 9 \cdot \frac{1}{9} (3 x + 1 - ln ∣ 3 x + 1 ∣)

Simplify

9 \cdot \frac{1}{9} (3 x + 1 - ln ∣ 3 x + 1 ∣) : 3 x + 1 - ln ∣ 3 x + 1 ∣

= 3 x + 1 - ln ∣ 3 x + 1 ∣

Add a constant to the solution

= 3 x + 1 - ln ∣ 3 x + 1 ∣ + C

f (x) = \frac{x}{x + 3}, at x = 4

\frac{1}{s ( s - 2 ) ^{2}}

y = \frac{x}{64 + x ^{2}}, (0, 0)

\frac{d}{d x} (2 sin (x) - 1)

\int (ln (x y)) d x

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