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

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

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

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

- \frac{1}{2} ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣ - \frac{3}{2} ln ∣ x + 3 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{2}{x + 2} d x = 2 ln ∣ x + 2 ∣

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{e ^{x}}{3 x})

\frac{\partial}{\partial x} (ln (x + y) + x cos (y))

\int_{0}^{3} \frac{7 t}{( t - 4 ) ^{2}} d t

7 x^{e} + e^{9 x}

f (x) = \frac{x ^{2}}{5 + x}

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