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

The integral of x/((x+1)(x+2)(x+3)^2) is -1/4 ln|x+1|+2ln|x+2|-7/4 ln|x+3|+3/(2(x+3))+C

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

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

- \frac{1}{4} ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣ - \frac{7}{4} ln ∣ x + 3 ∣ + \frac{3}{2 ( x + 3 )} + C

Solution steps

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

Take the partial fraction of

\frac{x}{( x + 1 ) ( x + 2 ) ( x + 3 ) ^{2}} : - \frac{1}{4 ( x + 1 )} + \frac{2}{x + 2} - \frac{7}{4 ( x + 3 )} - \frac{3}{2 ( x + 3 ) ^{2}}

= \int - \frac{1}{4 ( x + 1 )} + \frac{2}{x + 2} - \frac{7}{4 ( x + 3 )} - \frac{3}{2 ( x + 3 ) ^{2}} 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}{4 ( x + 1 )} d x + \int \frac{2}{x + 2} d x - \int \frac{7}{4 ( x + 3 )} d x - \int \frac{3}{2 ( x + 3 ) ^{2}} d x

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

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

\int \frac{7}{4 ( x + 3 )} d x = \frac{7}{4} ln ∣ x + 3 ∣

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

= - \frac{1}{4} ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣ - \frac{7}{4} ln ∣ x + 3 ∣ - (- \frac{3}{2 ( x + 3 )})

Simplify

= - \frac{1}{4} ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣ - \frac{7}{4} ln ∣ x + 3 ∣ + \frac{3}{2 ( x + 3 )}

Add a constant to the solution

= - \frac{1}{4} ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣ - \frac{7}{4} ln ∣ x + 3 ∣ + \frac{3}{2 ( x + 3 )} + C

What is the integral of x/((x+1)(x+2)(x+3)^2) ?
The integral of x/((x+1)(x+2)(x+3)^2) is -1/4 ln|x+1|+2ln|x+2|-7/4 ln|x+3|+3/(2(x+3))+C