What is the integral of (3-x)/(x^3+4x^2+3x) ?

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

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integral of (3-x)/(x^3+4x^2+3x)

\int \frac{3 - x}{x ^{3} + 4 x ^{2} + 3 x} d x

- 2 ln ∣ x + 1 ∣ + ln ∣ x + 3 ∣ + ln ∣ x ∣ + C

Solution steps

\int \frac{3 - x}{x ^{3} + 4 x ^{2} + 3 x} d x

Expand

\frac{3 - x}{x ^{3} + 4 x ^{2} + 3 x} : \frac{3}{x ^{3} + 4 x ^{2} + 3 x} - \frac{x}{x ^{3} + 4 x ^{2} + 3 x}

Apply the Sum Rule:

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

= \int \frac{3}{x ^{3} + 4 x ^{2} + 3 x} d x - \int \frac{x}{x ^{3} + 4 x ^{2} + 3 x} d x

\int \frac{3}{x ^{3} + 4 x ^{2} + 3 x} d x = ln ∣ x ∣ - \frac{3}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x + 3 ∣

\int \frac{x}{x ^{3} + 4 x ^{2} + 3 x} d x = \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x + 3 ∣

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

Simplify

ln ∣ x ∣ - \frac{3}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x + 3 ∣ - (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x + 3 ∣) : - 2 ln ∣ x + 1 ∣ + ln ∣ x + 3 ∣ + ln ∣ x ∣

= - 2 ln ∣ x + 1 ∣ + ln ∣ x + 3 ∣ + ln ∣ x ∣

Add a constant to the solution

= - 2 ln ∣ x + 1 ∣ + ln ∣ x + 3 ∣ + ln ∣ x ∣ + C

\frac{d}{d x} (4 ln (cos (x)))

x \to 3 - lim (2 x - 1)

\int_{1}^{4} 5 t ln (t) d t

x \to - 3 lim (- \frac{5}{( x + 3 ) ^{2}})

\frac{d}{d x} (\frac{8 - x e ^{x}}{x + e ^{x}})

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