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

The integral of (1-3x)/(3+2x) is 1/4 (-3(3+2x)+11ln|3+2x|)+C

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integral of (1-3x)/(3+2x)

\int \frac{1 - 3 x}{3 + 2 x} d x

\frac{1}{4} (- 3 (3 + 2 x) + 11 ln ∣ 3 + 2 x ∣) + C

Solution steps

\int \frac{1 - 3 x}{3 + 2 x} d x

Apply u-substitution

: \int \frac{- 3 u + 11}{4 u} d u

= \int \frac{- 3 u + 11}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{- 3 u + 11}{u} d u

Expand

\frac{- 3 u + 11}{u} : - 3 + \frac{11}{u}

= \frac{1}{4} \cdot \int - 3 + \frac{11}{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

= \frac{1}{4} (- \int 3 d u + \int \frac{11}{u} d u)

\int 3 d u = 3 u

\int \frac{11}{u} d u = 11 ln ∣ u ∣

= \frac{1}{4} (- 3 u + 11 ln ∣ u ∣)

Substitute back

u = 3 + 2 x

= \frac{1}{4} (- 3 (3 + 2 x) + 11 ln ∣ 3 + 2 x ∣)

Add a constant to the solution

= \frac{1}{4} (- 3 (3 + 2 x) + 11 ln ∣ 3 + 2 x ∣) + C

x \to 0 lim (\frac{x ^{3} - x}{x})

n = 1 \sum \infty ∣ sin (n) ∣

\frac{d}{d x} ((cos (x)) (ln (tan (x))))

(2 x + 3)^{\frac{1}{4}}

y^{2} + 8 x = 16, y^{2} - 24 x = 48

What is the integral of (1-3x)/(3+2x) ?
The integral of (1-3x)/(3+2x) is 1/4 (-3(3+2x)+11ln|3+2x|)+C