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

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

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

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

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

Solution steps

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

Expand

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

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

\int \frac{x ^{2}}{( x + 1 ) ( x + 2 ) ( x + 3 )} d x = \frac{1}{2} ln ∣ x + 1 ∣ - 4 ln ∣ x + 2 ∣ + \frac{9}{2} ln ∣ x + 3 ∣

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

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

= \frac{1}{2} ln ∣ x + 1 ∣ - 4 ln ∣ x + 2 ∣ + \frac{9}{2} ln ∣ x + 3 ∣ - ln ∣ x + 1 ∣ + 4 ln ∣ x + 2 ∣ - 3 ln ∣ x + 3 ∣ - (\frac{1}{2} ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x + 3 ∣)

Simplify

\frac{1}{2} ln ∣ x + 1 ∣ - 4 ln ∣ x + 2 ∣ + \frac{9}{2} ln ∣ x + 3 ∣ - ln ∣ x + 1 ∣ + 4 ln ∣ x + 2 ∣ - 3 ln ∣ x + 3 ∣ - (\frac{1}{2} ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x + 3 ∣) : - ln ∣ x + 1 ∣ + ln ∣ x + 3 ∣ + ln ∣ x + 2 ∣

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

Add a constant to the solution

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

f (x) = (4 x - 9)^{2}

y^{^{'}} - y = x, y (0) = 1

x \to - 2 lim (x^{3} + 8)

\int x (x - 1)^{4} d x

f (x) = 4^{x^{9}}

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