Integrate $\int{\frac{1}{(x+1)(x+2)^2(x+3)^3}dx}$

Question

How to integrate this

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

I tried to use that $$\int{\frac{1}{(x+1)(x+2)^2(x+3)^3}dx} = P_{1}(x)/Q_{1}(x) + \int{P_{2}(x)/Q_{2}(x)dx}$$ where $$\frac{P_{1}(x)}{Q_{1}(x)}=\frac{ax^2+bx+c}{(x+2)(x+3)^2}$$ and $$\frac{P_{2}(x)}{Q_{2}(x)}= \frac{dx^2+kx+m}{(x+1)(x+2)(x+3)}$$ but I got terrible coeff a,b,c,d,k,m. Is there an easier way?

Answer 1

Write $$\frac{1}{(x+1)(x+2)^2(x+3)^3}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}+\frac{D}{(x+3)}+\frac{E}{(x+3)^2}+\frac{F}{(x+3)^3}$$ Solve for $A,B,C,D,E,F$ by comparing coefficients and integrate each term separately.

Answer 2

Hint:

Write as

$$\frac{1}{(x+1)(x+2)^2(x+3)^3}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}+\frac{D}{(x+3)}+\frac{E}{(x+3)^2}+\frac{F}{(x+3)^3}$$

The partial fraction expansion is:

$$\frac{2}{x+2}-\frac{17}{8(x+3)}-\frac{1}{(x+2)^2}-\frac{5}{4(x+3)^2}-\frac{1}{2(x+3)^3}+\frac{1}{8(x+1)}$$

Answer 3

For an alternative approach write the required integral in the form $\int{\frac{1}{(x+1)(x+a)^2(x+b)^3}dx}$ then use $\frac{1}{(x+a)^2} = - \frac{d}{da} \frac{1}{x+a}$ and similarly for $\frac{1}{(x+b)^3}$ in terms of $\frac{d^2}{d b^2} \frac{1}{x+b}$, so that you have to first evaluate $I(a,b) = \int{\frac{1}{(x+1)(x+a)(x+b)}dx}$. The required integral is $\frac{d}{d a} \frac{d^2}{d b^2} I(a,b)$, finally setting $a = 2$, $b = 3$.